Question

the reaction 2Na + 2H0 -> 2NaOH + H2, how many grams of hydrogen are produced if 120. g of sodium and 80. g of water are available?

Answer 1

Answer: 4.48 grams of H2

Step-by-step explanation:

This is a problem of limiting reactants. We will use the given amounts of reactant and see which produces the least amount of H2. This will be the amount of H2 that can actually be produced.

Case 1: All sodium is used

120. g Na

x 1 mol Na / (22.99 g Na)

x 1 mol H2 / (2 mol Na)

x 2.016 g H2 / (1 mol H2)

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5.26 g H2 produced

Case 2: All water is used

80. g H2O

x 1 mol H2O / (18.016 g H2O)

x 1 mol H2 / (2 mol H2O)

x 2.016 g H2 / (1 mol H2)

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4.48 g of H2 produced

From these calculations, we see that we can only produce 4.48 g of H2 before we run out of water (if we had excess water, we could have produced 5.26 g of H2 with the amount of sodium we have).