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Explanation:

Q 1;

Given data:

Mass of Na₂CO₃ produced = ?

Mass of sodium oxide = 50.0 g

Solution:

Chemical equation:

2Na₂O + CO    →     Na₂CO₃ + 2Na

Number of moles of sodium oxide:

Number of moles = mass/ molar mass

Number of moles = 50 g/ 62 g/mol

Number of moles = 0.806 mol

Now we will compare the moles of sodium oxide with sodium carbonate.

                              Na₂O          :               Na₂CO₃

                                 2              :                   1

                             0.806            :              1/2×0.81 = 0.403

Mass of sodium carbonate:

Mass = number of moles × molar mass

Mass = 0.403 mol × 106 g/mol

Mass = 42.7 g

Q 2;

Given data:

Moles of Al₂(SO₄)₃ produced = ?

Mass of Al = 35 g

Solution:

Chemical equation:

2Al + 3H₂SO₄    →     Al₂(SO₄)₃ + 3H₂

Number of moles of Aluminium:

Number of moles = mass/ molar mass

Number of moles = 35 g/ 27 g/mol

Number of moles = 1.3 mol

Now we will compare the moles of sodium oxide with sodium carbonate.

                              Al               :            Al₂(SO₄)₃

                               2               :                   1

                             1.3               :              1/2×1.3 = 0.65

So from 35 gram of Al 0.65 moles of Al₂(SO₄)₃ are produced in the presence of excess sulfuric acid.