Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water. if 1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide, calculate the percent yield of sodium sulfate

The % yield is 27.0 %

Explanation:

Step 1: Data given

Mass of sulfuric acid = 4.9 grams

Mass of sodium hydroxide = 7.8 grams

Mass of sodium sulfate produced = 1.92 grams

Molar mass H2SO4 = 98.08 g/mol

Molar mass NaOH = 40 g/mol

Molar mass Na2SO4 = 142.04 g/mol

Step 2: The balanced equation

H2SO4 + 2NaOH → Na2SO4 + 2H2O

Step 3: Calculate moles H2SO4

Moles H2SO4 = Mass H2SO4 / Molar mass H2SO4

Moles H2SO4 = 4.9 grams / 98.08 g/mol =

Moles H2SO4 = 0.05 moles

Step 4: Calculate moles NaOH

Moles NaOH = 7.8 grams / 40 g/mol

Moles NaOH = 0.195 moles

Step 5: Calculate limiting reactant

For 1 mole H2SO4 consumed ,we need 2 moles NaOH to produce 1 mole Na2SO4 and 2 moles H2O

H2SO4 is the limiting reactant. It will completely be consumed (0.05 moles).

NaOH is in excess. There will react 2*0.05 = 0.1 moles

There will remain 0.195 -0.1 = 0.095 moles NaOH

Step 6: Calculate moles Na2SO4

For 1 mole H2SO4 consumed ,we need 2 moles NaOH to produce 1 mole Na2SO4

For 0.05 moles H2SO4, we have 0.05 moles Na2SO4

Step 7: Calculate mass of Na2SO4

Mass Na2SO4 = Moles Na2SO4 * Molar mass Na2SO4

Mass = 0.05 moles * 142.04 g/mol = 7.102

This is the theoretical yield

Step 8: Calculate the percent yield of Na2SO4

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

% yield = (1.92 /  7.102) *100% = 27.0 %

The % yield is 27.0 %

first one is the correct answer

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