Question

A reaction was conducted between barium nitrate and sodium phosphate. 3 Ba(NO3)2 + 2 Na3PO4 → 6 NaNO3 + Ba3(PO4)2 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

Answer:

The % yield is 95.0 %

Explanation:

Step 1: Data given

Number of moles Ba(NO3)2 = 0.3 moles

Number of moles Na3PO4 = 0.25 moles

Number of Ba3(PO4)2 = 0.095 moles

Step 2: The balanced equation

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

Step 3: Calculate the limiting reactant

For 3 moles Ba(NO3)2 we need 2 moles Na3PO4 to produce 6 moles NaNO3 and 1 mol Ba3(PO4)2

Ba(NO3)2 is the limiting reactant. It will compeltely be consumed completely (0.3 moles). Na3PO4 is in excess. There will react 0.20 moles.

There will remain 0.25-0.20= 0.05 moles

Step 4: Calculate moles Ba3(PO4)2

For 3 moles Ba(NO3)2 we need 2 moles Na3PO4 to produce 6 moles NaNO3 and 1 mol Ba3(PO4)2

For 0.3 moles Ba(NO3)2 we'll have 0.3/3 = 0.1 mol Ba3(PO4)2

Step 5: Calculate percent yield

% yield = (actual yield / theoretical yield) * 100%

% yield = (0.095 / 0.1) *100 %

% yield = 95.0 %

The % yield is 95.0 %

Answer 2

The percent yield of the reaction is calculated in terms of theoretical yield. The percent yield for reaction is 95%.

What is the limiting reactant in the reaction?

The reactant with the lesser concentration in the reaction as required is termed as the limiting reactant. The limiting reactant determines the yield of the product in the reaction.

The balanced chemical equation for the reaction is:

$\rm 3\;Ba(NO_3)_2\;+\;2\;Na_3PO_4\;\rightarrow\;6\;NaNO_3\;+\;Ba_3(PO_4)_2$

The complete reaction of 3 moles of barium nitrate requires 2 moles of sodium phosphate. The available moles of barium nitrate is 0.3 mol. The moles of sodium phosphate requires is:

$\rm 3\;mol\;Ba(NO_3)_2=2\;mol\;Na_3PO_4\\0.3\;mol\;Ba(NO_3)_2=\dfrac{2}{3}\;\times\;0.3\;mol\;Na_3PO_4\\ 0.3\;mol\;Ba(NO_3)_2=0.2\;mol\;Na_3PO_4$

The moles of sodium phosphate required is 0.2 mol. The available moles are 0.25 mol. Thus, sodium phosphate is the excess reactant and barium nitrate is the limiting reactant.

The theoretical yield of the reaction is given as:

$\rm 3\;mol\;Ba(NO_3)_2=1\;mol\;Ba_3(PO_4)_2\\0.3\;mol\;Ba(NO_3)_2=\dfrac{1}{3} \;\times0.3\;mol\;Ba_3(PO_4)_2\\\\0.3\;mol\;Ba(NO_3)_2=0.1\;mol\;Ba_3(PO_4)_2$

The theoretical yield of the reaction is 0.1 mol. The experimental yield of the reaction is 0.095 mol.

The percent yield of the reaction is given as:

$\rm Percent\;yield=\dfrac{Experimental\;yield}{Theoretical\;yield} \;\times\;100$

Substituting the values:

$\rm Percent\;yield=\dfrac{0.095}{0.1} \;\times\;100\\\\Percent\;yield=95\%$

The percent yield for the reaction is 95%.

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