Question

Suppose 3.5g of calcium chloride reacts with 6.39g of silver nitrate.

How many moles of calcium nitrate would be produced?

Balanced Equation =
Limiting Reactant =
Excess Reactant =

Answer 1

CaCl₂+2AgNO₃⇒Ca(NO₃)₂+2AgCl

mole CaCl₂: 3.5 g : 111 g/mole = 0.032

mole AgNO₃: 6.39 g : 170 g/mole = 0.038

mole: coefficient

CaCl₂: 0.032 : 1 = 0.032

AgNO₃: 0.038 : 2 = 0.019

AgNO₃ smaller= limiting reactant, then CaCl₂ = excess reactant