How many grams of sodium sulfate will be produced if 48.6 grams of sodium chloride reacted with excess sulfuric acid?

To determine the amount of sodium sulfate produced from the reaction of sodium chloride with sulfuric acid, we first need to consider the balanced chemical equation:

[ 2 \text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{HCl} ]

From the equation, 2 moles of NaCl produce 1 mole of Na2SO4. The molar mass of NaCl is approximately 58.44 g/mol, so 48.6 grams of NaCl is about 0.831 moles. This means 0.831 moles of NaCl will produce 0.4155 moles of Na2SO4. The molar mass of Na2SO4 is about 142.04 g/mol, so the total mass of sodium sulfate produced would be approximately 59.1 grams.