Question

The reaction of 2.9 g of H2SO4 with 1.7 g of NaOH produces 1.78 g of Na2SO4.

What is the percent yield of Na2SO4 and which was the limiting reagent?

Answer 1

To determine the percent yield of Na2SO4, we first identify H2SO4 as the limiting reagent based on stoichiometry, calculate the theoretical yield from the moles of H2SO4,

Step-by-step explanation:

To calculate the percent yield of Na2SO4, we first need to determine the theoretical yield based on the stoichiometry of the balanced chemical equation for the reaction between H2SO4 and NaOH, which is:

H2SO4 + 2 NaOH → Na2SO4 + 2 H2O

First, we calculate the moles of H2SO4 and NaOH provided using their respective molar masses. From the balanced equation, we see that 1 mole of H2SO4 reacts with 2 moles of NaOH, indicating that NaOH is in excess. Therefore, H2SO4 is the limiting reagent. We then convert the moles of H2SO4 to moles of Na2SO4 using the stoichiometry of the balanced equation, and from there, we determine the theoretical mass of Na2SO4.

The percent yield is calculated using the formula:

Percent Yield = (Actual Yield/Theoretical Yield) × 100%

Comparing the actual yield (1.78 g) to the theoretical yield, we can calculate the percent yield of Na2SO4. This would determine Na2SO4's efficiency in the reaction.