The actual amount of substance each required or produced by the reaction is obtained by stoichiometry.
From the reaction equation;
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
Number of moles in 1.56 grams of H2(g) = 1.56 grams/2.00g/mol = 0.78 moles
If 2 moles of water yields 1 mole of H2
x moles of water yields 0.78 moles of H2
x = 2 × 0.78/1 = 1.56 moles of water
Mass of water required = 1.56 moles of water × 18 g/mol = 28 grams of water
The statement that the reaction requires 27.9 grams of H2O is true.
Also;
The number of moles in 1.55g of NaOH = 1.55g/40g/mol = 0.039 moles
If 2 moles of water produces 2 moles of NaOH
1.56 moles of water produces 1.56 × 2/2 = 1.56 moles of NaOH
Mass of NaOH = 1.56 moles of NaOH * 40 g/mol = 62.4 g of NaOH
The statement that the reaction also produces 1.55 grams of NaOH is false.
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