Answer:
CuCl₂ is the excessive reactant and NaNO₃ the limiting reactant, 13.7g of NaCl can be formed.
Explanation:
Based on the reaction:
CuCl₂ + 2NaNO₃ → 2NaCl + Cu(NO₃)₂
To solve the problem we must convert the mass of each reactant in order to find limiting reactant as follows:
<em>Moles CuCl₂ -Molar mass: 134.45g/mol-:</em>
35g * (1mol / 134.45g) = 0.26 moles
<em>Moles NaNO₃ -Molar mass: 84.99g/mol-:</em>
20g * (1mol / 84.99g) = 0.235 moles
For a complete reaction of 0.235 moles of NaNO₃ there are required:
0.235 moles NaNO₃ * (1 mol CuCl₂ / 2 mol NaNO₃) = 0.118 moles CuCl₂
As there are 0.26 moles CuCl₂,
<h3>CuCl₂ is the excessive reactant and NaNO₃ the limiting reactant</h3><h3 />
As 2 moles of NaNO₃ produce 2 moles of NaCl, he moles of NaCl produced are: 0.235 moles NaCl. The mass is:
<em>Mass NaCl -Molar mass: 58.44g/mol-:</em>
0.235 moles NaCl * (58.44g / mol) =
<h3>13.7g of NaCl can be formed</h3>
