Silicon dioxide reacts vigorously with magnesium powder to form magnesium silicide, a dark blue solid - Leaving Cert Chemistry - Question c - 2015

To find the number of moles of magnesium silicide (Mg₂Si) produced from 7.6 g, we first determine its molar mass:

The molar mass of Mg₂Si:

• Magnesium (Mg): 24.3 g/mol, hence for 2 Mg: $2 imes 24.3 = 48.6$ g/mol
• Silicon (Si): 28.1 g/mol

Total molar mass of Mg₂Si = $48.6 + 28.1 = 76.7$ g/mol

Now, using the formula: ext{Number of moles} = rac{ ext{mass}}{ ext{molar mass}}

We can calculate:

ightarrow 0.099 ext{ mol} ext{ (approx. 0.10 mol)} $$ Since 1 mole of Mg₂Si reacts with 2 moles of Mg, the number of moles of magnesium required: - From the equation: 0.10 mol Mg₂Si requires $2 imes 0.10 = 0.20$ mol of Mg.

From the stoichiometry of the reaction:

• 1 mole of Mg₂Si reacts with 4 moles of HCl.

Thus: $ext{Number of moles of HCl} = 4 imes ext{moles of Mg₂Si}$

Substituting the number of moles of Mg₂Si calculated earlier:

• For 0.10 mol Mg₂Si:
  $ext{Number of moles of HCl} = 4 imes 0.10 = 0.40 ext{ mol}$

From the equation:

• 1 mole of Mg₂Si produces 2 moles of MgCl₂.

Thus for 0.10 mol of Mg₂Si: $ext{Moles of MgCl₂} = 2 imes 0.10 = 0.20 ext{ mol}$

Calculate mass:

• Molar mass of MgCl₂
  $= 24.3 + 2 imes 35.5 = 95.3 ext{ g/mol}$

$ext{Mass} = ext{Moles} imes ext{Molar mass} = 0.20 ext{ mol} imes 95.3 ext{ g/mol} = 19.06 ext{ g} ext{ (approximately 19 g)}$

From the reaction:

• 1 mole of SiH₄ reacts with 2 moles of O₂.

For the reaction:

• 0.10 mol of Mg₂Si produces 0.10 mol of SiH₄.

Thus: $ext{Moles of O₂ required} = 2 imes 0.10 = 0.20 ext{ mol}$

At room temperature and pressure, 1 mole of gas occupies 24 liters: $ext{Volume of O}_2 = 0.20 ext{ mol} imes 24 ext{ L/mol} = 4.8 ext{ L}$