Question

milk of magnesia, an aqueous suspension of magnesium hydroxide, is used as an antacid in the reaction below.
how many molecules of hcl would have to be present to form 58.54 g of mgcl₂?
mg(oh)₂(s) + 2hcl(aq) ⇌ 2h₂o(l) + mgcl₂(aq)

Explanation:

Step1: Calculate the molar mass of \(MgCl_2\)

The molar mass of \(MgCl_2\) is \(M = 24.31+(2\times35.45)=95.21\space g/mol\)

Step2: Calculate the number of moles of \(MgCl_2\)

Using the formula \(n=\frac{m}{M}\), where \(m = 58.54\space g\) and \(M = 95.21\space g/mol\). So \(n=\frac{58.54}{95.21}=0.615\space mol\)

Step3: Use the stoichiometry of the reaction

From the balanced equation \(Mg(OH)_2(s)+2HCl(aq)
ightleftharpoons2H_2O(l)+MgCl_2(aq)\), the mole ratio of \(HCl\) to \(MgCl_2\) is \(2:1\). So the number of moles of \(HCl\) is \(n_{HCl}=2\times n_{MgCl_2}=2\times0.615 = 1.23\space mol\)

Step4: Calculate the number of molecules of \(HCl\)

Using Avogadro's number \(N = n\times N_A\), where \(N_A=6.022\times 10^{23}\space mol^{-1}\). So \(N=1.23\times6.022\times 10^{23}=7.41\times 10^{23}\)

Answer:

\(7.41\times 10^{23}\) molecules of \(HCl\)