Answer: 6.162g of Ag2SO4 could be formed
Explanation:
Given;
0.255 moles of AgNO3
0.155 moles of H2SO4
Balanced equation will be given as;
2AgNO3(aq) + H2SO4(aq) -> Ag2SO4(s) + 2HNO3(aq)
Seeing that 2 moles of AgNO3 is required to react with 1 moles of H2SO4 to produce 1 mole of Ag2SO4,
Therefore the number of moles of Ag2SO4 produced is given by,
n(Ag2SO4) = 0.255 mol of AgNO3 ×
[0.155mol H2SO4 ÷ 2 mol AgNO3] x
[ 1 mol Ag2SO4 ÷ 1 mol H2SO4]
= 0.0198 mol of Ag2SO4.
mass = no of moles x molar mass
From literature, molar mass of Ag2SO4 = 311.799g/mol.
Thus,
Mass = 0.0198 x 311.799
= 6.162g
Therefore, 6.162g of Ag2SO4 could be formed
