Question

A quantity of 0.225 g of a metal M (molar mass = 27.0 g/mol) liberated 0.303 L of molecular hydrogen (measured at 17°C and 741 mmHg) from an excess of hydrochloric acid. Deduce from these data the corresponding equation and write formulas for the oxide and sulfate of M.

Answer 1

Answer:

Oxide of M is $M_2O_3$ and sulfate of $M_2(SO_4)_3$

Explanation:

0.303 L of molecular hydrogen gas measured at 17°C and 741 mmHg.

Let moles of hydrogen gas be n.

Temperature of the gas ,T= 17°C =290 K

Pressure of the gas ,P= 741 mmHg= 0.9633 atm

Volume occupied by gas , V = 0.303 L

Using an ideal gas equation:

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.9633 atm\times 0.303 L}{0.0821 atm L/mol K\times 290 K}=0.01225 mol$

Moles of hydrogen gas produced = 0.01225 mol

$2M+2xHCl\rightarrow 2MCl_x+xH_2$

Moles of metal =$\frac{0.225 g}{27.0 g/mol}=8.3333 mol$

So, 8.3333 mol of metal M gives 0.01225 mol of hydrogen gas.

$\frac{8.3333}{0.01225 mol}=\frac{2}{x}$

x = 2.9 ≈ 3

$2M+6HCl\rightarrow 2MCl_3+3H_2$

$MCl_3\rightarrow M^{3+}+Cl^-$

Formulas for the oxide and sulfate of M will be:

Oxide of M is $M_2O_3$ and sulfate of $M_2(SO_4)_3$.