Question

 Calculate the mass of the product of 6.40 g of magnesium with 1.32 g of oxygen . 

Answer 1

The balanced reaction equation is:

2Mg + O₂ $\rightarrow$ 2MgO.

As one can see from the equation, 2 moles of magnesium react with 1 mol of oxygen and 2 moles of magnesium oxide are formed:

$\frac{n(Mg)}{2} = n(O_2)=\frac{n(MgO)}{2}$ .

The number of the moles is the mass $m$ divided by the molar mass $M$ :

$n = \frac{m}{M}$ .

The molar masses of Mg, O₂ and MgO are 24.31, 32.00 and 40.30 g/mol, respectively. Therefore, the number of the moles of Mg are O₂ are:

$n(Mg) = \frac{6.40}{24.31} = 0.2633$ mol.

$n(O_2) = \frac{1.32}{32.00} = 0.04125$ mol.

The number of the moles of Mg divided by two 0.2633/2 = 0.1316 mol is higher than the number of the moles of O₂. Thus, Mg is in excess. Therefore, the quantity of the product, MgO, must be calculated from the number of the moles of O₂:

$n(MgO) = 2n(O_2) = 2·0.04125 = 0.0825$ mol.

Finally, the mass of magnesium oxide formed is:

$m(MgO) =n·M = 0.0825·40.30 = 3.32$ g.

Answer: the mass of the product of 6.40 g of magnesium with 1.32 g of oxygen is 3.32 g.