Question

b) If 35 grams of copper (II) chloride react with 20 grams of sodium nitrate, how much sodium chloride can be formed? what is the limiting reactant ? what is the excessive reactant ?

Answer 1

CuCl₂ is the excessive reactant and NaNO₃ the limiting reactant, 13.7g of NaCl can be formed.

Step-by-step 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:

Moles CuCl₂ -Molar mass: 134.45g/mol-:

35g * (1mol / 134.45g) = 0.26 moles

Moles NaNO₃ -Molar mass: 84.99g/mol-:

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,

CuCl₂ is the excessive reactant and NaNO₃ the limiting reactant

As 2 moles of NaNO₃ produce 2 moles of NaCl, he moles of NaCl produced are: 0.235 moles NaCl. The mass is:

Mass NaCl -Molar mass: 58.44g/mol-:

0.235 moles NaCl * (58.44g / mol) =

13.7g of NaCl can be formed