When 10 grams of sodium is added to 15 grams of water, 0.43 g of hydrogen gas is produced.
2Na + 2H2O → 2NaOH + H2
Number of moles of sodium = 10g/23g/mol = 0.43 moles
Number of moles of water = 10 g/18 g/mol = 0 55 moles
Since the reaction is 2:2, sodium is the limiting reactant
Now;
2 moles of sodium metal yields 1 mole of hydrogen gas
0.43 moles of sodium metal yields 0.43 moles × 1 mole/2 moles = 0.215 moles of hydrogen gas
Mass of hydrogen gas = Number of moles × molar mass = 0.215 moles × 2 g/mol = 0.43 g
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