The question is incomplete, the complete question is;
Using the following equation 2 NaOH(aq) + H2SO4(aq) → 2 H2O(aq) + Na2SO4(aq) how many grams of sodium sulfate will be formed if you start with 200 grams of sodium hydroxide and you have an excess of sulfuric acid
Answer:
355.1 g
Explanation:
The equation of the reaction is;
2 NaOH(aq) + H2SO4(aq) → 2 H2O(aq) + Na2SO4(aq)
We have been told that H2SO4 is in excess so NaOH is the limiting reactant. Therefore;
Number of moles in 200g of NaOH = 200g/40g/mol = 5 moles
So;
2 moles of NaOH yields 1 mole of Na2SO4
5 moles of NaOH will yield 5 * 1/2 = 2.5 moles of Na2SO4
Molar mass of Na2SO4 = 142.04 g/mol
Mass of Na2SO4= 2.5 moles * 142.04 g/mol = 355.1 g
