Question

Calculate the number of grams of magnesium chloride that could be obtained from
17.0 g of HCl when HCl is reacted with an excess of magnesium oxide.

Answer 1

Hint: Magnesium oxide (MgO) is produced when magnesium (Mg) burns at high temperatures, producing a bright, hot flame. It is difficult to measure the heat of reaction, or enthalpy, because the reaction is so rapid.

Complete step by step solution:
Magnesium oxide reacts exothermically with hydrochloric acid and produces magnesium chloride, liquid water, and heat. Heat is produced because the reaction is exothermic.
\[\begin{align}
& Mg+{}^{1}/{}_{2}{{O}_{2}}\to MgO \\
& MgO+2HCl\to MgC{{l}_{2}}+{{H}_{2}}O \\
\end{align}\]
We can see from the above reaction that 2 moles of HCl react with 1 mole of MgO to form 1 mole of \[MgC{{l}_{2}}\].
Molar mass of H is 1 g/mol.
Molar mass of Cl is 35.5 g/mol.
Molar mass of HCl = 1 g/ mol + 35.5 g/mol = 36.5 g/mol

Therefore, two moles of HCl will have a molar mass of =\[2\,mol\times 36.5g/mol=73g\]
Molar mass of Mg is 24.3 g/mol.
Molar mass of Cl is 35.5 g/mol.
Molar mass of \[MgC{{l}_{2}}=24.3\text{ }g/mol\text{ }+\text{ }\left( 2\times 35.5\text{ }g/mol \right)=95.2\text{ }g/mol\]
Therefore, one mole of \[MgC{{l}_{2}}\] will have a molar mass of = \[1\,mol\times 95.2g/mol=95.2g\]
Now rest is just simple unitary method calculations.

73g of HCl gives \[MgC{{l}_{2}}\] = 95.2g
1g of HCl will give \[MgC{{l}_{2}}\] = \[\dfrac{95.2}{73}\]g
17g of HCl will give \[MgC{{l}_{2}}\] = \[\dfrac{95.2}{73}\times 17g=22.16g\]

Therefore, the number of grams of magnesium chloride obtained is 22.16g.

Note: When solid Mg reacts with HCl (hydrochloric acid), magnesium chloride (\[MgC{{l}_{2}}\]), hydrogen gas, and heat are produced. This reaction can be given as:
\[Mg+HCl\to MgC{{l}_{2}}+{{H}_{2}}\]