Answer :
Explanation:
The balanced equation for the reaction is given as;
2NaOH + H2SO4 → Na2SO4 + 2H2O
a. How many moles of sodium sulfate are produced if 13 g of sulfuric acid is used in the reaction?
From the reaction;
1 mol of H2SO4 produces 1 mol of Na2SO4
Converting 13g of sulphuric acid to mol;
Number of moles = Mass / Molar mass
Number of moles = 13g / 98.0785 g/mol
Number of moles = 0.1325 mol
Since the mol ratio is 1: 1. This means
0.1325 mol of sodium sulfate would be formed.
b. How many molecules of water are produced if 2.0 g of sodium sulfate is produced in the above reaction?
From the reaction;
1 mol of Na2SO4 is produced alongside 2 mol of H2O
Converting 2g of sodium sulfate mol;
Number of moles = Mass / Molar mass
Number of moles = 2g / 142.04 g/mol
Number of moles = 0.01408 mol
Since the mol ratio is 1: 2. This means
0.02816 mol of water would be formed.
1 mol of water = 6.022×10^23 molecules
0.02816 mol = x
Solving for x;
x = 1.696 ×10^22 molecules
c. What mass of sulfuric acid is required to completely react with 3.4 x 1024 formula units of sodium hydroxide?
From the reaction;
1 mol of H2SO4 reacts with 2 mol of NaOH
1 mol = 6.022×10^23 units
x mol = 3.4 x 10^24 units
Solving for x;
x = 5.646 mol of NaOH
Since the mole ratio is 1:2;
5.646 mol / 2 = 2.823 mol of sulphuric acid
Converting too mass;
Mass = Number of moles * Molar mass
Mass = 2.823 mol * 98.0785 g/mol
Mass = 276.88 g