What volume of hydrogen gas is produced when 49.7 g of sodium reacts completely according to the following reaction at 25 °C and 1 atm?

Question

sodium (s) + water(l)sodium hydroxide (aq) + hydrogen(g)

Answer 1

The first thing you need to do is write out the reaction like this:

2 Na(s) + 2 H2O(l) —> 2NaOH(aq) + H2 (g)

If you have 49.7 grams of Na and excess water then the limiting reactant is Na, therefore we can convert from moles of Na to moles of H2 using the mole-to-mole conversion.

n = g / MM

49.7 / 23 = 2.16 moles Na

Now using the balanced reaction convert to moles of H2:

( 2.16 moles Na ) * (1 mole H2 / 2 moles Na) = 1.08 moles H2

Because H2 is a gas you can use the ideal gas law equation to solve for the volume at STP.

PV = nRT

(1)(V) = 1.08 * 0.0821 * 298

We have to convert from C to K by adding 273. The 0.0821 is the R constant.

Solving for the volume you get 26.42 L or 26,423 ml.