Start by writing the balanced chemical equation that describes this reaction
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.
Now, in order for all the moles of sodium to take part in the reaction, you need
As you can see, you have
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.
Finally, to convert this to grams, use the molar mass of hydrogen gas
The answer must be rounded to one significant figure, the number of sig figs you have for your values. Ryan J. Sodium reacts vigorously with water to produce hydrogen and sodium hydroxide 2Na(s) + 2H2O(i) -> 2NaOH(aq) + H2(g). If 0.027 g of sodium react with excess water, what volume of hydrogen at STP is formed? 1 Expert Answer 2Na(s) + 2H2O(l) ==> 2NaOH(aq) + H2(g) ... balanced equation 0.027g......xs........................................?......... moles Na used = 0.027 g x 1 mol Na/23 g = 0.00117 moles Na moles H2 produced = 0.00117 mol Na x 1 mol H2 / 2 mol Na = 0.000587 moles H2 formed At STP 1 mol of an idea gas = 22.4 L Volume H2 formed = 0000587 mol x 22.4 L/mole = 0.0131 L = 0.013 L (2 significant figures) |