Answer:
84.4g of AgCl
Step-by-step 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