How many moles of hydrogen gas are produced when 20 g of sodium metal react with 10 g of water?

How many moles of hydrogen gas are produced when 20 g of sodium metal react with 10 g of water?

Answer:
\(0.6 \mathrm{~g} \mathrm{H}_{2}\)

Explanation:
Start by writing the balanced chemical equation that describes this reaction

\(\mathrm{Na}_{(s)}+2 \mathrm{H}_{2} \mathrm{O}_{(l)} \rightarrow 2 \mathrm{NaOH}_{(a q)}+\mathrm{H}_{2(g)} \uparrow\)

Sodium and water react in a 1 : 2 mole ratio, so the next thing to do here is to convert the given masses to moles by using the molar mass of sodium and the molar mass of water, respectively.

Now, in order for all the moles of sodium to take part in the reaction, you need

This means that water will act as a limiting reagent, i.e. it will be completely consumed before all the moles of sodium will get the chance to take part in the reaction.

So, you can say that the reaction will consume 0.555 moles of water.

To find the number of moles of hydrogen gas produced, use the 2:1 mole ratio that exists between water and hydrogen gas.

The answer must be rounded to one significant figure, the number of sig figs you have for your values.