Answer to Question #149659 in General Chemistry for Bhawani Shankar

Question

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water. If 2.31 of water is produced from the reaction of 7.85 of sulfuric acid and 11.7 of sodium hydroxide, calculate the percent yield of water. Round your answer to 3 significant figures.

Answer 1

Equation for the reaction:

H₂SO₄ + 2NaOH $\to$ Na₂SO₄ + H₂O

Assuming that:

"The units for the values for quantities of the reagents is not given, I want to assume that;

2.31mL of water is produced from 7.85mL of H₂SO₄ and 11.7g of NaOH"

Determining the limiting reagent:

Moles of NaOH present in 11.7g

Moles of NaOH $=\dfrac{Mass}{MM}$

$=\dfrac{11.7g}{40g/mol} = 0.2925mol$

Moles of water in 2.31g produced ;

$Moles of water =\dfrac{2.31g}{18g/mol} = 0.128mol$

Actual moles of NaOH used to form water;

Mole ratio of water:NaOH = 1:2

Thus moles of NaOH use $=\dfrac{2}{1}x0.128 = 0.2567moles$

But there were a total of 0.2925 moles of NaOH present. Therefore, not all the NaOH present reacted, suggesting that H₂SO₄ could have been the limiting reagent.

Theoretical moles of water that would be produced if all the NaOH was reacted would be;

Theoretical Moles of H₂O $=\dfrac{1}{2}xmoles of NaOH present$

$=\dfrac{1}{2} x 0.2925mol = 0.14625mol$

$Percent yield = \dfrac{(actual yield)}{(thoretical)} x 100$

% Yield $=\dfrac{0.128}{0.14625} x 100$

$$

$= 87.5$ % yield