Chemistry: How many grams of iron are needed to react with 30.0 g of sulfur?

How many grams of iron are needed to react with, №13014787, 19.04.2020 05:24

Chemistry

4fecr2o7 + 8 k2co3 + o2 2 fe2o3 + 8 k2cro4 + 8 co2 how many grams of iron (ii) dichromate are required to produce 44.0 grams of carbon dioxide? how many grams of oxygen gas are required to produce 100.0 grams of ferric oxide? if 300.0 grams of iron (ii) dichromate react, how many grams of oxygen gas will be consumed? how many grams of iron (iii) oxide will be produced from 300.0 grams of ferrous dichromate?

Step-by-step answer

P Answered by Specialist

9

a)135.9 grams

b) 10 grams

c) 8.83g

d) 88.3g

Explanation:

the balanced reaction shows that four molecules of iron (II) dichromate will react with eight molecules of potassium carbonate and one molecule of oxygen to give two molecules of ferric oxide, eight molecules of potassium chormate and carbon dioxide each.

a) How many grams of iron (II) dichromate are required to produce 44.0 grams of carbon dioxide?

The molar mass of carbon dioxide = 44 g/mol

The molar mass of iron (II) dichromate = 271.8 g/mol

Thus as we need one mole of carbon dioxide we need to have half moles of iron (II) dichromate = 0.5 X 271.8 = 135.9 grams

b) How many grams of oxygen gas are required to produce 100.0 grams of ferric oxide?

The molar mass of ferric oxide =160 g/mol

the molar mass of oxygen = 32 g/mol

two moles of ferric oxide are obtained from one mole of oxygen

thus

320g of ferric oxide from 32 grams of oxygen

for 100g of ferric oxide the mass of oxygen required = 10 grams

c) If 300.0 grams of iron (II) dichromate react, how many grams of oxygen gas will be consumed?

The molar mass of iron (II) dichromate = 271.8 g/mol

the molar mass of oxygen = 32 g/mol

With four moles of iron (II) dichromate one mole of oxygen reacts

thus 4 X 271.8 grams of iron (II) dichromate reacts with = 32g of oxygen

hence 300 grams will react with =

$\frac{32X300}{4X271.8}=8.83g$

d) How many grams of iron (III) oxide will be produced from 300.0 grams of ferrous dichromate?

The molar mass of iron (II) dichromate = 271.8 g/mol

The molar mass of ferric oxide =160 g/mol

from 4 moles of iron dichromate two moles of ferric oxides are formed

thus

4 X 271.8 grams of iron dichromate will give = 2X160 g of ferric oxide

300 grams will give =

$\frac{2X160X300}{4X271.8}=88.3g$

Chemistry

4fecr2o7 + 8 k2co3 + o2 2 fe2o3 + 8 k2cro4 + 8 co2 how many grams of iron (ii) dichromate are required to produce 44.0 grams of carbon dioxide? how many grams of oxygen gas are required to produce 100.0 grams of ferric oxide? if 300.0 grams of iron (ii) dichromate react, how many grams of oxygen gas will be consumed? how many grams of iron (iii) oxide will be produced from 300.0 grams of ferrous dichromate?

Step-by-step answer

P Answered by Master

9

a)135.9 grams

b) 10 grams

c) 8.83g

d) 88.3g

Explanation:

the balanced reaction shows that four molecules of iron (II) dichromate will react with eight molecules of potassium carbonate and one molecule of oxygen to give two molecules of ferric oxide, eight molecules of potassium chormate and carbon dioxide each.

a) How many grams of iron (II) dichromate are required to produce 44.0 grams of carbon dioxide?

The molar mass of carbon dioxide = 44 g/mol

The molar mass of iron (II) dichromate = 271.8 g/mol

Thus as we need one mole of carbon dioxide we need to have half moles of iron (II) dichromate = 0.5 X 271.8 = 135.9 grams

b) How many grams of oxygen gas are required to produce 100.0 grams of ferric oxide?

The molar mass of ferric oxide =160 g/mol

the molar mass of oxygen = 32 g/mol

two moles of ferric oxide are obtained from one mole of oxygen

thus

320g of ferric oxide from 32 grams of oxygen

for 100g of ferric oxide the mass of oxygen required = 10 grams

c) If 300.0 grams of iron (II) dichromate react, how many grams of oxygen gas will be consumed?

The molar mass of iron (II) dichromate = 271.8 g/mol

the molar mass of oxygen = 32 g/mol

With four moles of iron (II) dichromate one mole of oxygen reacts

thus 4 X 271.8 grams of iron (II) dichromate reacts with = 32g of oxygen

hence 300 grams will react with =

$\frac{32X300}{4X271.8}=8.83g$

d) How many grams of iron (III) oxide will be produced from 300.0 grams of ferrous dichromate?

The molar mass of iron (II) dichromate = 271.8 g/mol

The molar mass of ferric oxide =160 g/mol

from 4 moles of iron dichromate two moles of ferric oxides are formed

thus

4 X 271.8 grams of iron dichromate will give = 2X160 g of ferric oxide

300 grams will give =

$\frac{2X160X300}{4X271.8}=88.3g$

Chemistry

A student wants to reclaim the iron from an 18.0-gram sample of iron(III) oxide, which occurs according to the reaction shown below: 2Fe2O3(s) — 4Fe(s) + 30,(g) How many grams of iron can be reclaimed?

Step-by-step answer

P Answered by PhD

3

12.6g of Fe.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2Fe2O3(s) —> 4Fe(s) + 3O2(g)

Next, we shall determine the mass of Fe2O3 that decomposed and the mass of Fe that is produced from the balanced equation.

This is illustrated below:

Mola mass of Fe2O3 = (56 x 2) + (16x3) = 160g/mol

Mass of Fe2O3 from the balanced equation = 2 x 160= 320g

Molar mass of Fe = 56g/mol

Mass of Fe from the balanced equation = 4 x 56 = 224g

From the balanced equation above,

320g of Fe2O3 decomposes to produce 224g of Fe.

Finall, we can obtain the mass of the Fe produced from the decomposition of 18g of Fe2O3 as follow:

From the balanced equation above,

320g of Fe2O3 decomposes to produce 224g of Fe.

Therefore 18g of Fe2O3 will decompose to produce = (18x224)/320 = 12.6g of Fe.

Therefore, 12.6g of Fe is obtained from 18g of Fe2O3.

Chemistry

A student wants to reclaim the iron from an 18.0-gram sample of iron(III) oxide, which occurs according to the reaction shown below: 2Fe2O3(s) — 4Fe(s) + 30,(g) How many grams of iron can be reclaimed?

Step-by-step answer

P Answered by PhD

3

12.6g of Fe.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2Fe2O3(s) —> 4Fe(s) + 3O2(g)

Next, we shall determine the mass of Fe2O3 that decomposed and the mass of Fe that is produced from the balanced equation.

This is illustrated below:

Mola mass of Fe2O3 = (56 x 2) + (16x3) = 160g/mol

Mass of Fe2O3 from the balanced equation = 2 x 160= 320g

Molar mass of Fe = 56g/mol

Mass of Fe from the balanced equation = 4 x 56 = 224g

From the balanced equation above,

320g of Fe2O3 decomposes to produce 224g of Fe.

Finall, we can obtain the mass of the Fe produced from the decomposition of 18g of Fe2O3 as follow:

From the balanced equation above,

320g of Fe2O3 decomposes to produce 224g of Fe.

Therefore 18g of Fe2O3 will decompose to produce = (18x224)/320 = 12.6g of Fe.

Therefore, 12.6g of Fe is obtained from 18g of Fe2O3.

Chemistry

PLEASE HELP QUICK Lab - Limiting Reactant and Percent Yield Data (Trial 1) (5 Points) Observations during Reaction: 3CuCl2(aq) + 2Al(s) → 3Cu(s) +2AlCl3(aq) Measured Mass (g) N/A N/A Molar Mass ( ) 134.45 26.98 N/A N/A Actual Moles (mol) N/A N/A Analysis: The limiting reactant is _ because . . . Data (Trial 2) (5 Points) Observations during Reaction: 3CuCl2(aq) + 2Al(s) → 3Cu(s) +2AlCl3(aq) Measured Mass (g) Filter paper alone Filter paper + Cu Cu alone N/A Molar Mass ( ) 134.45 26.98 63.55 N/A Actual Moles (mol) N/A Analysis: The limiting reactant

Step-by-step answer

P Answered by PhD

2

im stuck on it too unfortunatley

Explanation:

Chemistry

PLEASE HELP QUICK Lab - Limiting Reactant and Percent Yield Data (Trial 1) (5 Points) Observations during Reaction: 3CuCl2(aq) + 2Al(s) → 3Cu(s) +2AlCl3(aq) Measured Mass (g) N/A N/A Molar Mass ( ) 134.45 26.98 N/A N/A Actual Moles (mol) N/A N/A Analysis: The limiting reactant is _ because . . . Data (Trial 2) (5 Points) Observations during Reaction: 3CuCl2(aq) + 2Al(s) → 3Cu(s) +2AlCl3(aq) Measured Mass (g) Filter paper alone Filter paper + Cu Cu alone N/A Molar Mass ( ) 134.45 26.98 63.55 N/A Actual Moles (mol) N/A Analysis: The limiting reactant

Step-by-step answer

P Answered by PhD

2

im stuck on it too unfortunatley

Explanation:

Chemistry

One of the reactions in a blast furnace used to reduce iron is shown above. how many grams of fe2o3 are required to produce 15.5 g of fe if the reaction occurs in the presence of excess co? a.11.1 g b.22.1 g c.30.0 g d.44.2 g

Step-by-step answer

P Answered by PhD

11

Answer : The correct option is, (b) 22.1 g

Solution : Given,

Mass of iron = 15.5 g

Molar mass of iron = 56 g/mole

Molar mass of Fe_2O_3 = 160 g/mole

First we have to calculate the moles of iron.

$\text{Moles of Fe}=\frac{\text{Mass of Fe}}{\text{Molar mass of Fe}}=\frac{15.5g}{56g/mole}=0.276moles$

Now we have to calculate the moles of Fe_2O_3 .

The balanced reaction is,

$Fe_2O_3+3CO\rightarrow 2Fe+3CO_2$

From the balanced reaction, we conclude that

As, 2 moles of iron obtained from 1 mole of Fe_2O_3

So, 0.276 moles of iron obtained from $\frac{0.276}{2}=0.138$ mole of Fe_2O_3

Now we have to calculate the mass of Fe_2O_3

$\text{Mass of }Fe_2O_3=\text{Moles of }Fe_2O_3\times \text{Molar mass of }Fe_2O_3$

$\text{Mass of }Fe_2O_3=(0.138mole)\times (160g/mole)=22.08g=22.1g$

Therefore, the amount of Fe_2O_3 required are, 22.1 grams.

Chemistry

One of the reactions in a blast furnace used to reduce iron is shown above. how many grams of fe2o3 are required to produce 15.5 g of fe if the reaction occurs in the presence of excess co? a.11.1 g b.22.1 g c.30.0 g d.44.2 g

Step-by-step answer

P Answered by PhD

11

Answer : The correct option is, (b) 22.1 g

Solution : Given,

Mass of iron = 15.5 g

Molar mass of iron = 56 g/mole

Molar mass of Fe_2O_3 = 160 g/mole

First we have to calculate the moles of iron.

$\text{Moles of Fe}=\frac{\text{Mass of Fe}}{\text{Molar mass of Fe}}=\frac{15.5g}{56g/mole}=0.276moles$

Now we have to calculate the moles of Fe_2O_3 .

The balanced reaction is,

$Fe_2O_3+3CO\rightarrow 2Fe+3CO_2$

From the balanced reaction, we conclude that

As, 2 moles of iron obtained from 1 mole of Fe_2O_3

So, 0.276 moles of iron obtained from $\frac{0.276}{2}=0.138$ mole of Fe_2O_3

Now we have to calculate the mass of Fe_2O_3

$\text{Mass of }Fe_2O_3=\text{Moles of }Fe_2O_3\times \text{Molar mass of }Fe_2O_3$

$\text{Mass of }Fe_2O_3=(0.138mole)\times (160g/mole)=22.08g=22.1g$

Therefore, the amount of Fe_2O_3 required are, 22.1 grams.

Chemistry

ron (II) hydroxide, Fe(OH)2, is very insoluble in water. Hence, if it is produced during an aqueous reaction it will precipitate from solution. How many grams of sodium hydroxide (a strong base) must be added to 42.7 mL of a 0.305 M aqueous solution of iron (II) nitrate, in order to precipitate all of the iron as Fe(OH)2

Step-by-step answer

P Answered by Specialist

1.04g of NaOH must be added to precipitate all of the iron

Explanation:

Based on the chemical reaction:

Fe(NO₃)₂(aq) + 2NaOH(aq) → Fe(OH)₂(s) + 2NaNO₃(aq)

Where 1 mole of Iron (II) nitrate reacts with 2 moles of NaOH to produce 1 mole of Fe(OH)₂

To solve this question we need to find the moles of Fe(NO₃)₂ and using the chemical reaction we can find the moles of NaOH and its mass:

Moles Fe(NO₃)₂:

0.0427L * (0.305mol / L) = 0.0130moles of Fe(NO₃)₂

Moles NaOH

0.0130moles of Fe(NO₃)₂ * (2mol NaOH / 1mol Fe(NO₃)₂) =

0.0260 moles NaOH

Mass -Molar mass NaOH = 40g/mol-:

0.0260 moles NaOH * (40g / mol) =

1.04g of NaOH must be added to precipitate all of the iron

Chemistry

1. Calculate the number of moles of carbon dioxide formed when 40.0 moles of oxygen is consumed in the burning of propane. Note the equation is unbalanced. C3H8 + O2 --- CO2 + H2O a. 40.0 mol CO2 b. 42.0 mol CO2 c. 20.0 mol CO2 d. 24.0 mol CO2 2. Iron reacts with superheated steam to form hydrogen gas and the oxide Fe3O4. Calculate the number of moles of hydrogen produced by 10.0 g of iron and enough steam. 3Fe (s) + 4H2O (g) --- Fe3O4 (s) + 4H2 (g) a. 0.239 mol H2 b. 239 mol H2 c. 2.39 mol H2 d. 2239 mol H2 3. Iron (III) oxide, also known as rust

Step-by-step answer

P Answered by Specialist

2

1. D (24.0 moles CO2)

2. A (.239 moles H2)

Explanation:

1. First Balance the equation

1 C3H8 + 5 O2 ---> 3 CO2 + 4 H2O

Then set up a stoiciometric equation so that the moles of O2 cancel out

40mol O2 x $\frac{3 mole CO2}{5 mole O2}$ = 24.0 moles CO2

2. Set up a stoichiometric equation

10 grams Fe x $\frac{1 mole Fe}{55.85 grams Fe}$ x $\frac{4 mole H2}{3 mole Fe}$ = 0.239 moles H2

How many grams of iron are needed to react with 30.0 g of sulfur?

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