Question

A 47.7 g sample of Cu is added to 116 mL of 5.5 M HNO3. Assuming the following reaction is only one that occuts;

Cu(s) + HNO3 ----> Cu(NO3)2 (aq) + H2O(ı) + NO(g)

Will the Cu react completely? What is the limiting reagent and what is the remaining compound in grams? How many grams of copper(II) nitrate will be obtained?

Answer 1

3 Cu (s) + 8 HNO₃ (aq) → 3 Cu(NO₃)₂ (aq) + 2 NO (g) + 4 H₂O (l)

Moles of Cu$=\frac{47.7} {63.546}=0.75moles$

Moles of HNO3 $=5.5×\frac{116}{1000}=0.638$

No

$HNO_3$ Is the limiting reagent

Moles of Cu(NO₃)₂$=0.638×\frac{3}{8}=0.239moles$

Cu(NO₃)₂ Molar mass: 187.56 g/mol

Mass $0.239×187.56=44.83grams$