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

Start by writing the balanced chemical equation that describes this reaction

#"Na"_ ((s)) + color(blue)(2)"H"_ 2"O"_ ((l)) -> 2"NaOH"_ ((aq)) + "H"_ (2(g)) uarr#

Sodium and water react in a #1:color(blue)(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.

#20 color(red)(cancel(color(black)("g"))) * ("1 mole Na")/(23.0color(red)(cancel(color(black)("g")))) = "0.870 moles Na"#

#10 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "0.555 moles H"_2"O"#

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

#0.870 color(red)(cancel(color(black)("moles Na"))) * (color(blue)(2)color(white)(.)"moles H"_2"O")/(1color(red)(cancel(color(black)("mole Na")))) = "1.74 moles H"_2"O"#

As you can see, you have

#overbrace("0.555 moles H"_2"O")^(color(purple)("what is available")) " "< " "overbrace("1.74 moles H"_2"O")^(color(purple)("what is needed"))#

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 #color(blue)(2):1# mole ratio that exists between water and hydrogen gas.

#0.555 color(red)(cancel(color(black)("moles H"_2"O"))) * "1 mole H"_2/(color(blue)(2)color(red)(cancel(color(black)("moles H"_2"O")))) = "0.2775 moles H"_2#

Finally, to convert this to grams, use the molar mass of hydrogen gas

#0.2775 color(red)(cancel(color(black)("moles H"_2))) * "2.016 g"/(1color(red)(cancel(color(black)("mole H"_2)))) = color(darkgreen)(ul(color(black)("0.6 g")))#

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