Aquantity 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.

Oxide of M is and sulfate of

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:

Moles of hydrogen gas produced = 0.01225 mol

Moles of metal =

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

x = 2.9 ≈ 3

Formulas for the oxide and sulfate of M will be:

Oxide of M is and sulfate of .