Question

When ammonia is mixed with hydrogen chloride (HCl) , the white solid ammonium chloride $N{H}_{4}Cl$ in produced. Suppose 10.0g ammonia is mixed with the same mass of hydrogen chloride. what substance will be present after the reaction has gone to completion and what will these masses be?

Short answer

The given an amount of ammonia mixed with mass of hydrogen chloride. one needs to find out the substance which will be present after the reaction has gone to completion and the masses of the compounds involved.

Step-by-step solution

Calculation of molar masses of NH4Cl, NH3,HCl

It is given that 10.0g of ammonia reacts with same mass of hydrogen chloride. one knows that molar masses of N, H and Cl are 14.007g/mol,1.0079g/mol and 35.453g/mol respectively.

Hence, one can determine the molar masses of $N{H}_{4}Cl,N{H}_3,HCl$as follows:

width="481">MolarmassofNH4Cl=14.007+4×1.0079+35.453g=53.50gMolarmassofNH3=14.007+3×1.0079=17.031gMolarmassofHCl=1.0079+35.453g=36.461g

Calculation of moles of NH3 in HCl 10.0g

We know 1 mol of $N{H}_3$ contains 17. 031g.

Hence

$1gofN{H}_{3}contains=\frac{1}{17.0337}mol$

$$\begin{gathered} 10gofN{H}_{3}contains=\frac{10.0}{17.0337}mol=0.578mol \\ 1gofHClcontains=\frac{1}{36.461}mol \\ 10gofHClcontains=\frac{10}{36.461}mol=0.274mol \end{gathered}$$

$$\begin{gathered} 1gofN{H}_{3}contains=\frac{1}{17.0337}mol \\ 10gofN{H}_{3}contains=\frac{10.0}{17.0337}mol=0.578mol \end{gathered}$$

$$\begin{gathered} 1moleofHClcontains=36.461g \\ 1gofHClcontains=\frac{1}{36.461}mol \\ 10gofHClcontains=\frac{10}{36.461}mol=0.274mol \end{gathered}$$

Calculating number of moles of NH4Cl and NH3

The balanced equation for the reaction is

$N{H}_3+HCl\to N{H}_{4}Cl$

From the equation, 1 mol reacts with 1 mol of HCl to give the 1 mol of $N{H}_{4}Cl$; hence after completion of reaction, the number of moles of $N{H}_{4}Cl$present in 0.274 mol and number of moles of $N{H}_3$is given by

$$\begin{gathered} \left(0.587\right)-\left(0.274\right)mol \\ =0.313mol \end{gathered}$$