01.08.2022

Zn + 2MnO2 + H2O → Zn(OH)2 + Mn2O3 Determine the limiting reactant if 25g of Zn and 30g of MnO2 are used and what mass of Zn(OH)2 will be produced?

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09.07.2023, solved by verified expert
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Balanced equation provided:

Zn + 2MnO2 + H2O → Zn(OH)2 + Mn2O3 Determine, №18010553, 01.08.2022 09:51

Also provided, 25 g of Zn and 30 g of MnO₂ are used

Find moles of Zn:

Zn + 2MnO2 + H2O → Zn(OH)2 + Mn2O3 Determine, №18010553, 01.08.2022 09:51

Zn + 2MnO2 + H2O → Zn(OH)2 + Mn2O3 Determine, №18010553, 01.08.2022 09:51

Zn + 2MnO2 + H2O → Zn(OH)2 + Mn2O3 Determine, №18010553, 01.08.2022 09:51

Find moles of MnO₂:

Zn + 2MnO2 + H2O → Zn(OH)2 + Mn2O3 Determine, №18010553, 01.08.2022 09:51

Zn + 2MnO2 + H2O → Zn(OH)2 + Mn2O3 Determine, №18010553, 01.08.2022 09:51

Zn + 2MnO2 + H2O → Zn(OH)2 + Mn2O3 Determine, №18010553, 01.08.2022 09:51

Therefore, MnO₂ is the limiting reactant.

Mr of Zn(OH)₂ is 99

using molar ratio find moles of Zn(OH)₂

2MnO₂ : Zn(OH)₂

2 : 1

0.345 : 0.1725

Zn(OH)₂ has 0.1725 moles.

Find mass of Zn(OH)₂

Zn + 2MnO2 + H2O → Zn(OH)2 + Mn2O3 Determine, №18010553, 01.08.2022 09:51

Zn + 2MnO2 + H2O → Zn(OH)2 + Mn2O3 Determine, №18010553, 01.08.2022 09:51

Zn + 2MnO2 + H2O → Zn(OH)2 + Mn2O3 Determine, №18010553, 01.08.2022 09:51

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Chemistry
Step-by-step answer
P Answered by Master

Balanced equation provided:

\sf Zn + 2MnO_2 + H_2O \rightarrow  Zn(OH)_2 + Mn_2O_3

Also provided, 25 g of Zn and 30 g of MnO₂ are used

Find moles of Zn:

\sf moles = \frac{mass}{Mr}

\sf moles = \frac{25}{65}

\sf moles = 0.3846 \ mol

Find moles of MnO₂:

\sf moles = \frac{mass}{Mr}

\sf moles = \frac{30}{87}

\sf moles = 0.345 \ mol

Therefore, MnO₂ is the limiting reactant.

Mr of Zn(OH)₂ is 99

using molar ratio find moles of Zn(OH)₂

2MnO₂ : Zn(OH)₂

2 : 1

0.345 : 0.1725

Zn(OH)₂ has 0.1725 moles.

Find mass of Zn(OH)₂

\sf mass = moles * Mr

\sf mass =  0.1725*99

\sf mass  = 17.069 \ g

Chemistry
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P Answered by PhD
Answer: chalcogens.
Explanation: Strontium is an alkaline earth metal, it always exhibits a degree of oxidation in its compounds +2.
Chalcogens are a group of 6 chemical elements (oxygen O, sulfur S, selenium se, tellurium te, polonium Po) that have an oxidation state of -2 => Chalcogens will combine with strontium in a ratio of 1:1.
Chemistry
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P Answered by PhD

Answer:

52.6 gram

Step-by-step explanation:

It is clear by the equation 2(27+3×35.5)= 267 gm of AlCl3 reacts with 6× 80 = 480 gm of Br2 . So 29.2 gm reacts = 480× 29.2/267= 52.6 gm

Chemistry
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P Answered by Master

Calcium (Ca)(On the periodic table, ionization energy increases as you go up and to the right of the periodic table)

Calcium (Ca)(On the periodic table, ionization energy increases as you go up and to the right of the
Chemistry
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P Answered by PhD

glycoproteins

Explanation:

A positive reaction for Molisch's test is given by almost all carbohydrates (exceptions include tetroses & trioses). It can be noted that even some glycoproteins and nucleic acids give positive results for this test (since they tend to undergo hydrolysis when exposed to strong mineral acids and form monosaccharides).

Chemistry
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P Answered by PhD

Answer:

Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.

An ideal gas is a theoretical gas that is considered to be composed of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P×V = n×R×T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. The universal constant of ideal gases R has the same value for all gaseous substances.

Explanation:

In this case, you know:

P= 0.884 atm

V= ?

n= Answer:Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N 0.857 moles (where 28 g/mole is the molar mass of N₂, that is, the amount of mass that the substance contains in one mole.)

R=0.082Answer:Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N

T= 328 K

Replacing in the ideal gas law:

0.884 atm×V= 0.857 moles× 0.082Answer:Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N ×328 K

Solving:

Answer:Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N

V= 26.07 L

The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.

Chemistry
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P Answered by PhD
15 moles.Explanation:Hello,In this case, the undergoing chemical reaction is:Clearly, since carbon and oxygen are in a 1:1 molar ratio, 15 moles of carbon will completely react with 15 moles of oxygen, therefore 15 moles of oxygen remain as leftovers. In such a way, since carbon and carbon dioxide are also in a 1:1 molar ratio, the theoretical yield of carbon dioxide is 15 moles based on the stoichiometry:Best regards.
Chemistry
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P Answered by PhD
Answer: 7.8125 g
Explanation: Given:
Original amount (N₀) = 500 g
Number of half-lives (n) = 9612/1602 = 6
Amount remaining (N) = ?
N = 1/2ⁿ × N₀
N = 1/2^6 × 500
N = 0.015625 × 500
N = 7.8125 g
Chemistry
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P Answered by PhD
Answer: a. basic
b. basic
c. acidic
d. neutral

Explanation: Acids and bases can be classified in terms of hydrogen ions or hydroxide ions, or in terms of electron pairs. (look at the picture)
Let us note that from the pH scale, a pH of;
0 - 6.9 is acidic
7 is neutral
8 - 14 is basic

But pH= - log [H^+]
pOH = -log [OH^-]
Then;
pH + pOH = 14
Hence;
pH = 14 - pOH

a. [H+] = 6.0 x 10-10M
pH= 9.22 is basic
b. [OH-] = 30 × 10-2M
pH = 13.5 is basic
C. IH+1 = 20× 10-7M
pH = 0.56 is acidic
d. [OH-] = 1.0 x 10-7M
pH = 7 is neutral
Answer: a. basic
b. basic
c. acidic
d. neutral

Explanation: Acids and bases can be classified
Chemistry
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P Answered by PhD
Answer: 306 L
Explanation: Using the ideal gas law,
PV = nRT
where R = 0.08206 L•atm/(mol•°K), solving for n gives
n = PV/(RT)
n = (845 mmHg) (270 L) / ((0.08206 L•atm/(mol•°K)) (24 °C))

Convert the given temperature to °K and the given pressure to atm:
24 °C = (273.15 + 24) °K ≈ 297.2 °K
(845 mmHg) × (1/760 atm/mmHg) ≈ 1.11 atm

Then the balloon contains
n = (1.11 atm) (270 L) / ((0.08206 L•atm/(mol•°K)) (297.2 °K))
n ≈ 12.3 mol
of He.

Solve the same equation for V :
V = nRT/P

Convert the target temperature to °K:
-50 °C = (273.15 - 50) °K = 223.15 °K

Then the volume under the new set of conditions is
V = (12.3 mol) (0.08206 L•atm/(mol•°K)) (223.15 °K) / (0.735 atm)
V ≈ 306 L

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