1CuCl2 + 2Na(NO3) --> 1Cu(NO3)2 + 2NaCl If 15 grams of copper(II) chloride react with 20 grams of sodium nitrate how much sodium chloride will be produced? g NaCl

13.45 grams of NaCl will be produced

Explanation:

Given:

Mass of CuCl2 = 15 grams

Molar mass of CuCl2 = 134.5 g/mol

Mass of Na(NO3) = 20grams

Molar mass of Na(NO3) = 85 g/mol

To find:

Amount of Sodium chloride produced = ?

Solution:

No of moles of CuCl2 = 15/134.5 = 0.11 moles

No of moles of Na(NO3) = 20/85 = 0.23 moles,

1CuCl2 + 2Na(NO3) --> 1Cu(NO3)2 + 2NaCl

from the reaction, 1 mole of copper chloride reacts with 2 mole of NaNO3 to produce 2 moles of sodium chloride.

Hence, 0.11 mole of copper chloride will react with 0.23 mole of NaNO3 to produce 0.23 moles of sodium chloride.

Mass of 0.23 moles of NaCl = 0.23 × M.M of NaCl

(Molar mass of NaCl = 58.5)

Mass of 0.23 moles of NaCl = 0.23 × 58.5 = 13.45 grams

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