What is the theoretical yield of sodium oxide (Na₂O) in grams when 20.0g of calcium oxide (CaO) reacts with an excess amount of Sodium chloride?

2NaCl+CaO->CaCl2+Na2O

The 2 at the end of the equation is not a coefficient it’s for Na2.
10.0g
22.1g
32.9g
5.6g

22.14g of Na₂O

Explanation:

Equation of reaction

2NaCl + CaO → CaCl₂ + Na₂O

To solve this problem, we'll employ mole concept and see how we can relate 20g of CaO to give us the theoretical yield of Na₂O.

Molar mass of CaO = (40 + 16) = 56g/mol

Molar mass of Na₂O = (2*23) + 16 = 62g/mol

From the equation of reaction,

1 mole of CaO = 1 mole of Na₂O

56g of CaO = 62g of Na₂O (use number of moles relationship with mass I.e number of moles = mass / molar mass to find the mass).

56g of CaO = 62g of Na₂O

20g of CaO = y g of Na₂O

y = (20 * 62) / 56

y = 22.14g of Na₂O

Therefore, 20g of CaO will produce 22.14g of Na₂O.

a corroding iron, if that's one of your choices