Chemistry percent yield

Question

4. In an experiment, 170.9 g of C_{2}*H_{4} was reacted with 52.75L of , 64.1 g of C*O_{2} is produced .What is the limiting reagent? ) What is the percent yield of this reaction? What is the limiting reagent? What is the percent yield?

Answer 1

To determine the limiting reagent and percent yield, we need to first write and balance the chemical equation for the reaction:

C₂H₄ + O₂ → CO₂ + H₂O

From the given information, we know that:

1. The mass of C₂H₄ = 170.9 g
2. The volume of O₂ = 52.75 L
3. The mass of CO₂ produced = 64.1 g

To find the limiting reagent, we need to calculate the amount (in moles) of each reactant and compare their ratios to the stoichiometric ratio in the balanced equation.

The amount of C₂H₄ can be calculated using its molar mass:

170.9 g C₂H₄ / (2 x 12.01 g/mol + 4 x 1.01 g/mol) = 3.58 mol C₂H₄

The amount of O₂ can be calculated using the ideal gas law:

n(O₂) = PV/RT = (1 atm)(52.75 L)/(0.0821 L atm/mol K)(298 K) = 2.16 mol O₂

Now we can compare the mole ratios of the reactants to the stoichiometric ratio in the balanced equation:

C₂H₄ : O₂ = 3.58 mol : 2.16 mol = 1.66 : 1

Stoichiometric ratio: C₂H₄ : O₂ = 1 : 3

Since the ratio of C₂H₄ to O₂ is less than the stoichiometric ratio, C₂H₄ is the limiting reagent. This means that all of the O₂ will be used up before all of the C₂H₄ is consumed.

To determine the theoretical yield of CO₂, we can use the mole ratio between C₂H₄ and CO₂ in the balanced equation:

1 mol C₂H₄ produces 1 mol CO₂

Therefore, the theoretical yield of CO₂ is:

3.58 mol C₂H₄ × 1 mol CO₂/1 mol C₂H₄ × 44.01 g/mol = 157.3 g CO₂

The percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100%:

Percent yield = (actual yield/theoretical yield) × 100%

The actual yield of CO₂ is given as 64.1 g. Therefore, the percent yield is:

Percent yield = (64.1 g/157.3 g) × 100% = 40.7%

Therefore, the limiting reagent is C₂H₄, and the percent yield of the reaction is 40.7%.