Aqueous sulfuric acid (H2SO4) reacts with solid sodium hydroxide (NaOH) to produce aqeous sodium sulfate (Na2SO4) and liquid water (H2O). what is the theoretical yield of water formed from the reaction of 5.9 g of sulfuric acid and 6.6 g of sodium hydroxide? Be sure your answer has the correct number of significant digits in it.

The theoretical yield of water formed is 2.2 grams

Explanation:

Step 1: Data given

Mass of H2SO4 = 5.9 grams

Mass of NaOH = 6.6 grams

Molar mass H2SO4 = 98.08 g/mol

Molar mass of NaOH = 40.0 g/mol

Step 2: The balanced equation

2NaOH + H2SO4 → Na2SO4 + 2H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles H2SO4 = 5.9 grams / 98.08 g/mol

Moles H2SO4 = 0.060 moles

Moles NaOH = 6.6 grams / 40.0 g/mol

Moles NaOH = 0.165 moles

Step 4: Calculate the limiting reactant

For 2 moles NaOH we need 1 mol H2SO4 to produce 1 mol Na2SO4 and 2 moles H2O

H2SO4 is the limiting reactant. It will completely be consumed (0.060 moles). NaOH is in excess. There will react 2*0.060 = 0.120 moles

There will remain 0.165 - 0.120 = 0.045 moles NaOH

Step 5: Calculate moles H2O

For 2 moles NaOH we need 1 mol H2SO4 to produce 1 mol Na2SO4 and 2 moles H2O

For 0.0600 moles H2SO4 we'll have 2*0.0600 = 0.120 moles H2O

Step 6: Calculate mass H2O

Mass H2O = 0.120 moles * 18.02 g/mol

Mass H2O = 2.16 grams

The theoretical yield of water formed is 2.2 grams