Question

In an experiment between 26.4 grams of Mg and O,, Mgo is produced. The percent

Answer 1

Answer:

39.6 g

Explanation:

The equation of the reaction is;

2Mg(s) + O2(g) --------> 2MgO(s)

To obtain the limiting reactant;

Number of moles in 26.4 g of Mg = 26.4g/24 g/mol = 1.1 moles

If 2 moles of Mg yields 2 moles of MgO

1.1 moles of Mg yields 1.1 * 2/2 = 1.1 moles of MgO

Number of moles in 26.4 g of O2 = 26.4 g/32g/mol = 0.825 moles

If 1 mole of O2 yields 2 moles of MgO

0.825 moles of O2 yields 0.825 moles * 2/1 = 1.65 moles of MgO

Hence Mg is the limiting reactant.

Theoretical yield of MgO = 1.1 moles of MgO * 40 g/mol = 44 g

Percent yield = 90%

Percent yield = actual yield/theoretical yield * 100

Actual yield = Percent yield * theoretical yield/100

Actual yield = 90 * 44/100

Actual yield = 39.6 g