Question

q.6 for the reaction of magnesium and hcl, if you react 25 grams of magnesium with 37 grams of hcl, which reactant is the limiting reactant and how much mgcl₂ will be produced by the reaction?

Explanation:

Step1: Write the balanced chemical equation

$$Mg + 2HCl ightarrow MgCl_2 + H_2$$

Step2: Calculate the molar mass of reactants

Molar mass of \(Mg\): \(M_{Mg}=24.31\space g/mol\)
Molar mass of \(HCl\): \(M_{HCl}=1.01 + 35.45=36.46\space g/mol\)

Step3: Calculate the number of moles of reactants

Moles of \(Mg\): \(n_{Mg}=\frac{25\space g}{24.31\space g/mol}\approx1.03\space mol\)
Moles of \(HCl\): \(n_{HCl}=\frac{37\space g}{36.46\space g/mol}\approx1.015\space mol\)

Step4: Determine the limiting reactant

From the balanced equation, mole ratio \(Mg:HCl = 1:2\)
If \(Mg\) is the limiting reactant, moles of \(HCl\) needed \(=2\times n_{Mg}=2\times1.03 = 2.06\space mol\). But we have only \(1.015\space mol\) of \(HCl\).
If \(HCl\) is the limiting reactant, moles of \(Mg\) needed \(=\frac{n_{HCl}}{2}=\frac{1.015}{2}=0.5075\space mol\). We have \(1.03\space mol\) of \(Mg\). So \(HCl\) is the limiting reactant.

Step5: Calculate moles of \(MgCl_2\) produced

From the balanced equation, mole ratio \(HCl:MgCl_2=2:1\)
Moles of \(MgCl_2\): \(n_{MgCl_2}=\frac{n_{HCl}}{2}=\frac{1.015}{2}=0.5075\space mol\)

Step6: Calculate mass of \(MgCl_2\) produced

Molar mass of \(MgCl_2\): \(M_{MgCl_2}=24.31+(2\times35.45)=95.21\space g/mol\)
Mass of \(MgCl_2\): \(m_{MgCl_2}=n_{MgCl_2}\times M_{MgCl_2}=0.5075\times95.21\approx48.3\space g\)

Answer:

The limiting reactant is \(HCl\) and the mass of \(MgCl_2\) produced is approximately \(48.3\space g\)