Chemistry: What is the volume of 50.2 g CO2 has at STP

What is the volume of 50.2 g CO2 has at STP, №9593549, 22.06.2020 07:10

Chemistry

At stp how many liters of oxygen are required to react completely with 4.5 liters of hydrogen to form water? 2h2(g)+o2(g)— 2h2o(g) 1.8 l 3.0 l 9.0 l 2.0 l how many moles of aluminum are needed to react completely with 4.9 mol of feo? 2al(s)+3feo(s)— 3fe(s)*al2o3(s) 4.9 mol of al 0.8 mol of al 3.3 mol of al 7.4 mol of al you are required to use the molar mass in all of the following stoichiometric calculations except volume-volume problems mass-mass problems mass-particle problems mass-volume problems iron(lll) oxide is formed when iron combines

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1

Do you have the answers i need help

Chemistry

1. at stp, how many liters of oxygen are required to react completely with 3.6 liters of hydrogen to form water? 2h2(g)+o2(g) - 2h2o(g) a. 1.8 l b. 3.6 l c. 2.0 l d. 2.4 l 2. which conversion factor do you use first to calculate the number of grams of co2 produced by the reaction of 50.6 g of ch4 with o2? the equation for the complete combustion of methane is: ch4(g) +2o2(g) - co2 + 2h2o(l) a. 1 mol ch4/ 16.0g ch4 b. 2 mol o2/1 mol co2 c. 16.0g ch4/1 mol co4 d. 44.0g co2/2 mol co2 3. which of the following statements is true about the following

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1. is a
2.is a
3.is b
hope this helps

Chemistry

At stp how many liters of oxygen are required to react completely with 4.5 liters of hydrogen to form water? 2h2(g)+o2(g)— 2h2o(g) 1.8 l 3.0 l 9.0 l 2.0 l how many moles of aluminum are needed to react completely with 4.9 mol of feo? 2al(s)+3feo(s)— 3fe(s)*al2o3(s) 4.9 mol of al 0.8 mol of al 3.3 mol of al 7.4 mol of al you are required to use the molar mass in all of the following stoichiometric calculations except volume-volume problems mass-mass problems mass-particle problems mass-volume problems iron(lll) oxide is formed when iron combines

Step-by-step answer

P Answered by PhD

1

Do you have the answers i need help

Chemistry

1. at stp, how many liters of oxygen are required to react completely with 3.6 liters of hydrogen to form water? 2h2(g)+o2(g) - 2h2o(g) a. 1.8 l b. 3.6 l c. 2.0 l d. 2.4 l 2. which conversion factor do you use first to calculate the number of grams of co2 produced by the reaction of 50.6 g of ch4 with o2? the equation for the complete combustion of methane is: ch4(g) +2o2(g) - co2 + 2h2o(l) a. 1 mol ch4/ 16.0g ch4 b. 2 mol o2/1 mol co2 c. 16.0g ch4/1 mol co4 d. 44.0g co2/2 mol co2 3. which of the following statements is true about the following

Step-by-step answer

P Answered by PhD

1. is a
2.is a
3.is b
hope this helps

Chemistry

Which conversion factor should you use first to calculate the number of grams of co2 produced by the reaction of 50.6 g of ch4 with o2? the complete combustion of methane is described by this equation: ch4(g) + 2o2(g) → co2(g) + 2h2o(l) 44.0 g co2 / 2 mol co2 16.0 g ch4 / 1 mol co4 1 mol ch4 / 16.0 g ch4 2 mol o2 / 1 mol co2

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P Answered by PhD

2

139.18 g.

Explanation:

For the balanced equation:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l),

It is clear that 1 mol of CH₄ react with 2 mol of O₂ to produce 1 mol of CO₂ and 2 mol of H₂O.

Firstly, we need to calculate the no. of moles of 50.6 g of CH₄:

no. of moles of CH₄ = mass/molar mass = (50.6 g)/(16.0 g/mol) = 3.1625 mol.

Using cross-multiplication:

1.0 mol of CH₄(g) produces → 1.0 mol of CO₂, from stichiometry.

∴ 3.1625 mol of CH₄(g) produces → 3.1625 mol of CO₂.

∴ The no. of grams of CO₂ produced = (no. of moles of CO₂)(molar mass of CO₂) = (3.1625 mol)(44.01 g/mol) = 139.18 g.

Chemistry

Which conversion factor should you use first to calculate the number of grams of co2 produced by the reaction of 50.6 g of ch4 with o2? the complete combustion of methane is described by this equation: ch4(g) + 2o2(g) → co2(g) + 2h2o(l) 44.0 g co2 / 2 mol co2 16.0 g ch4 / 1 mol co4 1 mol ch4 / 16.0 g ch4 2 mol o2 / 1 mol co2

Step-by-step answer

P Answered by PhD

2

139.18 g.

Explanation:

For the balanced equation:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l),

It is clear that 1 mol of CH₄ react with 2 mol of O₂ to produce 1 mol of CO₂ and 2 mol of H₂O.

Firstly, we need to calculate the no. of moles of 50.6 g of CH₄:

no. of moles of CH₄ = mass/molar mass = (50.6 g)/(16.0 g/mol) = 3.1625 mol.

Using cross-multiplication:

1.0 mol of CH₄(g) produces → 1.0 mol of CO₂, from stichiometry.

∴ 3.1625 mol of CH₄(g) produces → 3.1625 mol of CO₂.

∴ The no. of grams of CO₂ produced = (no. of moles of CO₂)(molar mass of CO₂) = (3.1625 mol)(44.01 g/mol) = 139.18 g.

Chemistry

If 4.50 g of hcl are reacted with 15.00 g of caco, according to the following balanced chemical equation, calculate the theoretical yield of.co. 2hcl + caco3 cacl2 + h2o + co2 if 2.50 g of co2 are isolated, after carrying out the above reaction, calculate the percent yield of co2.

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HCl is the limiting reactant and the theoretical yield is 2.72 g of CO2. If the actual yield was 2.50 g then, the percent yield is 92.0% when rounding off is done only for the final answer.

Further Explanation:

In order to determine the theoretical yield and the percent yield of CO2, the following steps must be done:

Determine the limiting reactant. This is the reactant that will determine the amount of CO2 that will actually form. Determine the theoretical yield for CO2 when the limiting reactant is used.Get the percent yield by getting the ratio of the actual yield stated in the problem and the calculated theoretical yield multiplied by 100.

Determining the Limiting Reactant

The Limiting Reactant (LR) will produce fewer moles of the products. To check which of the reactants HCl or CaCO3 is the LR, we do dimensional analysis:

For HCl:

$moles\ CO_{2}\ = (4.50\ g\ HCl)\(\frac{1\ mol\ HCl}{36.46094\ g})( \frac{1\ mol\ CO_{2} }{2\ mol\ HCl}) \\moles\ CO_{2}\ =\ 0.0617098$

For CaCO3:

$moles\ of\ CO_{2}\ = (15.00\ g\ CaCO_{3})\ (\frac{1\ mol\ CaCO_{3} }{100.0869\ g\ CaCO_{3} })\ (\frac{1\ mol\ CO_{2} }{1\ mol\ CaCO_{3} })\\moles\ of\ CO_{2}\ = \ 0.1499$

Since HCl produces fewer moles of CO2, then it is the limiting reactant. We will use the given amount to determine the theoretical yield for CO2.

Determining the Theoretical Yield

From Step 1, we know that 0.0617098 moles of CO2 will be produced. We will just convert this to grams.

$grams\ CO_{2}\ =\ (0.0617098\ mol\ CO_{2}) (\frac{44.01\ g\ CO_{2}}{1\ mol\ CO_{2}})\\grams\ CO_{2}\ =\ 2.71585$

Since the answer only requires 3 significant figures, the final answer is 2.72 grams CO2.

Determining the Percent Yield

Dividing the actual yield by the theoretical yield will give us the percent yield, which is an indicator of how efficient the experiment or the method used was.

From the problem, the actual yield was 2.50 g, hence, the percent yield is:

$percent\ yield\ of\ CO_{2}\ = (\frac{2.50\ g}{2.71585\ g}) (100)\\percent\ yield\ of\ CO_{2}\ = 92.05221$

Rounding off to three significant figures, the percent yield is 92.0%. This suggests that the method used is somewhat efficient in producing CO2.

Learn More

Learn More about Limiting Reactant link Learn More about Excess Reactant link Learn More about Stoichiometry link

Keywords: stoichiometry, theoretical yield, actual yield

Chemistry

For each pair below, select the sample that contains the largest number of moles. pair a 2.50 g o2 a.n.s: 2.50 g n2 pair b 21.5 g n2 a.n.s: 15.2 g co2 pair c 0.054 g o2 a.n.s: 0.081 co2

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9

Explanation:

Pair A:

nitrogen

Moles of oxygen: