Question

When 16.3 g of magnesium and 4.52 g of oxygen gas react, how many grams of magnesium oxide will be formed? Identify the limiting and excess reactants.

Answer 1

Answer:

22.77 g.

he limiting reactant is O₂, and the excess reactant is Mg.

Explanation:

• From the balanced reaction:

Mg + 1/2O₂ → MgO,

1.0 mole of Mg reacts with 0.5 mole of oxygen to produce 1.0 mole of MgO.

• We need to calculate the no. of moles of (16.3 g) of Mg and (4.52 g) of oxygen:

no. of moles of Mg = mass/molar mass = (16.3 g)/(24.3 g/mol) = 0.6708 mol.

no. of moles of O₂ = mass/molar mass = (4.52 g)/(16.0 g/mol) = 0.2825 mol.

So. 0.565 mol of Mg reacts completely with (0.2825 mol) of O₂.

∴ The limiting reactant is O₂, and the excess reactant is Mg (0.6708 - 0.565 = 0.1058 mol).

Using cross multiplication:

1.0 mole of Mg produce → 1.0 mol of MgO.

∴ 0.565 mol of Mg produce → 0.565 mol of MgO.

∴ The amount of MgO produced = no. of moles x molar mass = (0.565 mol)(40.3 g/mol) = 22.77 g.

Answer 2

Answer:

$m_{MgO}=11.3gMgO$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2Mg+O_2\rightarrow 2MgO$

Hence, for given amounts of magnesium and oxygen, one computes the available moles of magnesium with its given mass and atomic mass and the moles of magnesium that will completely react with 4.52 g of oxygen by using their 2:1 molar ratio in the chemical reaction in order to identify the limiting reactant:

$n_{Mg}^{available}=16.3gMg*\frac{1molMg}{24gMg}=0.679molMg\\\\n_{Mg}^{reacted}=4.52gO_2*\frac{1molO_2}{32gO_2}*\frac{2molMg}{1molO_2} =0.2825molMg$

In such a way, as just 0.2825 moles of magnesium react, magnesium is identified as the excess reactant and the oxygen the limiting reactant. Furthermore, with the reacting moles, one computes the yielded grams of magnesium oxide as:

$m_{MgO}=0.2825molMg*\frac{2molMgO}{2molMg}*\frac{40gMgO}{1molMgO} \\\\m_{MgO}=11.3gMgO$

Best regards.