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You have 2.2 mol Xe and 2.0 mol F₂, but when you carry out the reaction you end up with only 0.25 mol XeF₄. What is the percent yield of this experiment? Xe(g) + 2 F₂ (g) → XeF₄ (g)

Respuesta :

jbferrado

Answer:

The correct answer is 25 %

Explanation:

According to the chemical reaction:

Xe(g) + 2 F₂ (g) → XeF₄ (g)

1 mol of Xe(g) reacts with 2 mol of F₂(g), so the stoichiometric ratio os reactants is 2 mol F₂/mol Xe.

We have 2.2 mol Xe and 2.0 mol F₂, so the ratio is:

2.0 mol F₂/2.2 mol Xe = 0.909 mol F₂/mol Xe

The molar ratio of reactant we have is lower than the required, so F₂ is the limiting reactant.

By using the limiting reactant, we calculate the theoretical amount of product (XeF₄). For this, we know that 1 mol of XeF₄ is formed from 2 mol of F₂ (1 mol XeF₄/ 2 mol F₂), and we have 2.0 mol F₂:

2.0 mol F₂ x (1 mol XeF₄/ 2 mol F₂)= 1 mol XeF₄

If we only obtained 0.25 mol XeF₄, the percent yield of the experiment is:

Yield = experimental amount/theoretical amount x 100%

        = 0.25 mol XeF₄/ 1 mol XeF₄ x 100% = 25 %

pstnonsonjoku

The percent yield of the solution is 25%.

The equation of the reaction is;

Xe(g) + 2F₂ (g) → XeF₄ (g)

From the reaction equation;

1 mole of Xe reacts with 2 moles of F2

2.2 moles of Xe will react with 2.2 moles × 2 moles /2 moles

= 4.4 moles of F2

We can see that there is not enough  F2 hence it is the limiting reactant

2 moles of F2 yields 1 mole of XeF4

2 moles of F2 yields 2 moles ×  1 mole/2 moles = 1 mole of XeF4

% yield = actual yield/Theoretical yield × 100

% yield =  0.25 mole/ 1mole × 100

% yield =  25%

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