1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# Calcium hydroxide reacts with ammonium chloride to give ammonia, according to the following equation:$\mathrm{Ca}{\left(\mathrm{OH}\right)}_{2}+2{\mathrm{NH}}_{4}\mathrm{Cl}\to {\mathrm{CaCl}}_{2}+2{\mathrm{NH}}_{3}+2{\mathrm{H}}_{2}\mathrm{O}$If $5.35\mathrm{g}$ of ammonium chloride is used, calculate the weight of calcium chloride is formed.$\left[\mathrm{H}=1,\mathrm{N}=14,\mathrm{O}=16,\mathrm{Cl}=35.5,\mathrm{Ca}=40\right]$

Open in App
Solution

## Step 1: Given informationCalcium hydroxide $\left(\mathrm{Ca}{\left(\mathrm{OH}\right)}_{2}\right)$ reacts with ammonium chloride $\left({\mathrm{NH}}_{4}\mathrm{Cl}\right)$ to give ammonia $\left({\mathrm{NH}}_{3}\right)$, calcium chloride $\left({\mathrm{CaCl}}_{2}\right)$ and water $\left({\mathrm{H}}_{2}\mathrm{O}\right)$.The given reaction is as follows: $\mathrm{Ca}{\left(\mathrm{OH}\right)}_{2}+2{\mathrm{NH}}_{4}\mathrm{Cl}\to {\mathrm{CaCl}}_{2}+2{\mathrm{NH}}_{3}+2{\mathrm{H}}_{2}\mathrm{O}$Given mass of Ammonium chloride $\left({\mathrm{NH}}_{4}\mathrm{Cl}\right)=5.35\mathrm{g}$Step 2: Calculation of mass of ammonium chloride and calcium chloride$\begin{array}{rcl}\mathrm{Mass}\mathrm{of}2\mathrm{mol}{\mathrm{NH}}_{4}\mathrm{Cl}& =& 2×\left(14\mathrm{g}+1\mathrm{g}×4+35.5\mathrm{g}\right)\\ & =& 107\mathrm{g}\end{array}$$\begin{array}{rcl}\mathrm{Mass}\mathrm{of}1\mathrm{mol}{\mathrm{CaCl}}_{2}& =& 40\mathrm{g}+35.5\mathrm{g}×2\\ & =& 111\mathrm{g}\end{array}$Step 3: Calculation of required weight of calcium chloride from stoichiometric calculation$107\mathrm{g}$ of ammonium chloride produces $111\mathrm{g}$ of calcium chloride.Thus, $5.35\mathrm{g}$ of ammonium chloride will produce $\frac{111}{107}×5.35=5.55\mathrm{g}$ of calcium chloride.Therefore, $5.55\mathrm{g}$ of calcium chloride will be formed.

Suggest Corrections
1
Join BYJU'S Learning Program
Related Videos
Alkanes - Preparation
CHEMISTRY
Watch in App
Explore more
Join BYJU'S Learning Program