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.
$$\begin{array}{rl}& Mg+{}^1/{}_2{O}_2\to MgO\\ & MgO+2HCl\to MgC{l}_2+H_{2}O\end{array}$$
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 =$$2mol\times 36.5g/mol=73g$$
Molar mass of Mg is 24.3 g/mol.
Molar mass of Cl is 35.5 g/mol.
Molar mass of $$MgC{l}_2=24.3g/mol+(2\times 35.5g/mol)=95.2g/mol$$
Therefore, one mole of $$MgC{l}_2$$ will have a molar mass of = $$1mol\times 95.2g/mol=95.2g$$
Now rest is just simple unitary method calculations.

73g of HCl gives $$MgC{l}_2$$ = 95.2g
1g of HCl will give $$MgC{l}_2$$ = $${\displaystyle \frac{95.2}{73}}$$g
17g of HCl will give $$MgC{l}_2$$ = $${\displaystyle \frac{95.2}{73}}\times 17g=22.16g$$

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:
$$Mg+HCl\to MgC{l}_2+H_2$$