Question

AgNO3 + BaCl2 -> AgCl + Ba(NO3)2

Blanace the equation. If I carried out this reaction in the lab and produced 410.8 grams of Ba(NO3)2, how much silver nitrate AgNO3 did I start with?

Answer 1

molar mass of Ba(NO3)2 = 261.3 g/mol
number of mol of Ba(NO3)2 = (mass of Ba(NO3)2)/(molar mass of Ba(NO3)2)
= 410.8/261.3
= 1.572 mol

balance reaction is
2AgNO3 + BaCl2 -> 2AgCl + Ba(NO3)2

according to reaction taking place
1 mol of Ba(NO3)2 required 2 mol of AgNO3
1.572 mol of Ba(NO3)2 required (2*1.572) mol of AgNO3
1.572 mol of Ba(NO3)2 required 3.144 mol of AgNO3
so,
number of mol of AgNO3 required = 3.144 mol

molar mass of AgNO3 = 169.9 g/mol
mass of AgNO3 = (number of mol of AgNO3 required)*(molar mass of AgNO3)
= 3.144*169.9
= 534.2 g

Answer: 534.2 g