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
