When silver nitrate reacts with barium chloride,silver chloride and barium nitrate are formed.How many grams of silver chloride are formed when 10.8 g of silver nitrate reacts with 15.0 g of barium chloride?

Question

Quizplus · Accepted Answer

1) Write the balanced chemical equation:
AgNO3 + BaCl2 → AgCl + Ba(NO3)2
2) Determine the limiting reactant:
AgNO3 → molar mass = 169.87 g/mol 
10.8 g ÷ 169.87 g/mol = 0.0636 mol
BaCl2 → molar mass = 208.23 g/mol
15.0 g ÷ 208.23 g/mol = 0.0721 mol
Thus, AgNO3 is the limiting reactant because it produces less moles of product.
3) Calculate the mass of AgCl produced:
Molar ratio of AgNO3 to AgCl is 1:1
0.0636 mol AgNO3 × 1 mol AgCl × 143.32 g/mol AgCl = 9.11 g AgCl
Therefore, the answer is A) 9.11 g of silver chloride are formed.

When Silver Nitrate Reacts with Barium Chloride,silver Chloride and Barium