Gaseous butane (CH3(CH2)2 CH3) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O).

Question

If 18.5g of water is produced from the reaction of 34.3g of butane and 87.1g of oxygen gas, calculate the percent yield of water. Round your answer to 3 significant figures.

= _____ %

Answer 1

Steps:

1) Find smaller theoretical yield of water

2) Use percent yield equation.

Balanced Equation:

2C4H10+13O2→8CO2+10H2O

34.4g C4H10 (1mol butane / 58.12 g/mol butane) (10 mol water / 2 mol butane) (18.02g water / 1 mol water)=53.328 g water

87.1 g O2 (1 mol O2 / 32.00 g O2) (10 mol water / 13 mol O2) (18.02g water / 1 mol water)=37.729g water

Smaller theoretical yield of water is 37.729 g water, which comes from O2, so O2 is the limiting reactant.

Real theoretical yield = smaller theoretical yield = 37.729 g water

%yield=(^{experimental yield} / _{theoretical yield})100

%yield=(^{18.5g water}/_{37.729g water})100

%yield=49.0% yield water

Answer 2

To find percent yield you will need the formula: %yield= actual yield/ theoretical yield x 100.

Since we are given the actual yield, we need to find the theoretical yield.

In order to calculate the theoretical yield, we need to determine if either butane or O₂ is the limiting reactant.

We will start off by writing out the balanced equation:

2C₄H₁₀ + 13O₂ -------> 8CO₂ + 10H₂O

Balancing our equation will help us find the mole ratio between the reactants and products. This reaction tells us 2 moles of butane and 13 moles of oxygen gas are used to create 10 moles of water and 8 moles of carbon dioxide.

Next, we need to find our limiting reactant. (Our limiting reactant will tell us which reactant gets used up first, thus which reactant will determine how much product can be formed).

If we are given 34.3g of butane and 87.1g of O₂ both values will need to be converted to moles.

Molar Mass of Butane: 58g

Molar Mass of O₂: 32g

34.3g butane x 1 mole/ 58g = 0.591 moles of butane

87.1g O₂ x 1 mole/ 32g = 2.72 moles of O₂

You can start with either reactant to determine limiting reactant, I am going to try butane first. We will need to multiply the amount of moles we are given by the stoichiometric coefficients to determine if we have enough of the second reactant for the first reactant to be used up. Simply put "If we have ___ of reactant 1, how much of reactant 2 is needed".

0.591 moles of butane x (13 moles of O₂/ 2 moles of butane) = 3.8415 moles of O₂ needed. Since, we need more O₂ than was given in the question, it can be stated that O₂ is our limiting reactant.

To be sure, we will repeat the previous step for O₂.

2.72 moles of O₂ x (2 moles of butane/ 13 moles of O₂) = 0.418 moles of butane.

Since we have more moles than needed of the butane, we can say that O₂ is in fact the limiting reactant.

Since all the moles of O₂ will be used up, we can use the molar coefficients and given values to determine the theoretical yield of water in the above reaction.

2.72 moles of O₂ x (10 moles of H₂O / 13 moles of O₂) = 2.09 moles of H₂O is our theoretical yield.

From here we can either convert the theoretical yield to grams or the actual yield to moles (both will work out to the same answer).

I am going to convert my actual yield to moles.

18.5 g H₂O x (1 mole H₂O/ 18 g H₂O) = 1.03 moles of H₂O is our actual yield.

Plug these values into our percent yield equation:

% yield = actual yield/ theoretical yield x 100

= 1.03/ 2.09 x 100= 49.3% is our percent yield of water.