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In an experiment, 8.50 g of methane, CH4, reacted with 15.9 g of oxygen gas, as shown in the following equation: CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) Determine the percent yield if 9.77 g of carbon dioxide was obtained

Respuesta :

The balanced chemical reaction is written as:
CH4(g) + 2 O2(g) = CO2(g) + 2 H2O(g)
We use this reaction to determine the theoretical yield of carbon dioxide from the given reactants. We do as follows:

8.50 g CH4 ( 1 mol / 16.05 g ) = 0.53 mol CH4
15.9 g O2 ( 1 mol / 32 g ) = 0.50 mol O2 <----------LIMITING REACTANT

0.50 mol O2 ( 1 mol CO2 / 2 mol O2 ) ( 44.01 g / 1 mol ) = 11.00 g CO2 theoretical

% yield = 9.77 g / 11.00 g x 100= 88.80%

Answer:

Percentage Yield  = 89.39 %

Explanation:

First thing's first, we have to write down the balanced chemical equation;

CH4(g) + 2 O2(g) --> CO2(g) + 2 H2O(g)

It is crucial we identify the limiting reactant before proceeding. The limiting reactant tells us which reactant is the reaction dependent on and how much of the product can be obtained. We calculate it using the following steps;

1. Calculate the number of moles of each reactant present

Methane - Number of moles = mass / molar mass = 8.50 / 16 =  0.53125 mol

Oxygen - Number of moles = mass / molar mass = 15.9 / 32 =  0.4969 mol

2. Divide the actual number of moles of each reactant by its stoichiometric coefficient in the balanced chemical equation:

Methane - 0.53125 / 1 = 0.53125

Oxygen - 0.4969 / 2 = 0.24845

3. The reactant with the smallest mole ratio is limiting. The limiting reactant is oxygen.

Since the product we are interested in is the Carbon dioxide, From the equation, we have;

2 mol of O2 reacts to produce 1 mol of CO2.

Expressing in terms of mass; Mass = Number of moles * Molar mass

64g ( 2 * 32) = 44g ( 1* 44)

15.9 g = x

Upon solving for x;

x = (15.9 * 44) / 64

x = 10.93

Theoretical Yield = 10.93g

Practical Yield = 9.77g

Percentage Yield = (Practical Yield / Theoretical Yield) * 100

Percentage Yield = ( 9.77 / 10.93 ) * 100

Percentage Yield = 89.39%

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