2. 2AgNO3 + CaCl2 → 2AgCl + Ca (NO3) 2

How much AgCl could be obtained starting with 100g of each reactant.

84.4g of AgCl

Explanation:

Based on the reaction:

2AgNO₃ + CaCl₂ → 2AgCl + Ca (NO₃) ₂

2 moles of AgNO₃ and 1 mole of CaCl₂ priduce 2 moles of AgCl and 1 mole of Ca (NO₃) ₂

100g of each reactant are:

AgNO₃: 100g * (1mol / 169.87g) = 0.589 moles

CaCl₂: 100g * (1mol / 110.98g) = 0.901 moles

For a complete reaction of 0.901 moles of CaCl₂ are necessaries 0.901*2 = 1.802 moles of AgNO₃. As there are just 0.589moles, AgNO₃ is limitng reactant

0.589 moles of AgNO₃ produce:

0.589 moles * (2 moles AgCl / 2 moles AgNO₃) =

0.589 moles of AgCl. In mass:

0.589 moles of AgCl * (143.32g / mol) = 84.4g of AgCl