Question

What is the percent yield if 0.3 mol ba(no3)2 and 0.25 mol na3po4 react to produce 0.095 mol ba3(po4)2?

Answer 1

The chemical reaction is expressed as:

3Ba(NO3)2 + 2Na3PO4 = Ba3(PO4)2 + 6NaNO3

To determine the percent yield, we need to determine the theoretical yield of the reaction from the given amounts of the reactants. We do as follows:

0.3 mol 3Ba(NO3)2 ( 2 mol Na3PO4 / 3 mol Ba(NO3)2) = 0.2 mol Na3PO4

Therefore, the limiting reactant would be Ba(NO3)2 since it is consumed completely in the reaction.


Theoretical yield = 0.3 mol 3Ba(NO3)2 ( 1 mol Ba3(PO4)2 / 3 mol Ba(NO3)2) = 0.1 mol Ba3(PO4)2

Percent yield = actual yield / theoretical yield = 0.095 mol Ba3(PO4)2 / 0.1 mol Ba3(PO4)2 x 100 = 95% 

Answer 2

The percent yield if 0.3 mol Ba(NO₃)₂ and 0.25 mol Na₃PO₄ react to produce 0.095 mol Ba₃(PO₄)₂ is 95%.

How do we calculate percent yield?

Percent yield of any reaction will be calculated by using the below equation as:

% yield = (Actual yield / Theoretical yield) × 100%

Given balanced chemical reaction is:

3Ba(NO₃)₂ + 2Na₃PO₄ → Ba₃(PO₄)₂ + 6NaNO₃

From the stoichiometry of the reaction:

0.3 moles of Ba(NO₃)₂ = reacts with 2/3×0.3 = 0.2 moles of Na₃PO₄

Therefore, the limiting reactant would be Ba(NO₃)₂ since it is consumed completely in the reaction.

0.3 moles of Ba(NO₃)₂ = produces 1/3×0.3 = 0.1 moles of Ba₃(PO₄)₂

Given actual yield of Ba₃(PO₄)₂ = 0.095 mol

& theoretical yield = 0.1 mol

On putting values on the % yield equation, we get

% yield = (0.095/0.1) × 100% = 95%

Hence required percent yield is 95%.

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