If #"2300.0 g"# of #"Na"(s)# reacts in a sufficient amount of water, what is the theoretical yield of #"H"_2(g)# in grams?
2 Answers
Sodium metal
#1. H-OH(l) => Hcdot(g) + cancel(cdotOH(aq))#
#2. Nacdot(s) + cancel(cdotOH(aq)) => NaOH(aq)#
#1.# The#H-OH# bond is split in two. This is a violent step, and in real life there's bubbling in the water!
#2.# #Nacdot# then bonds with#cdotOH# .
Overall, we get the product of a (radical) single-replacement reaction:
#color(green)(Nacdot(s) + H_2O(l) -> NaOH(aq) + Hcdot(g))# or
#color(green)(2Na(s) + 2H_2O(l) -> 2NaOH(aq) + H_2(g))#
The first obvious piece of information is that your limiting reagent is sodium metal (since the water is in excess), so you can use that as the starting compound for your "mole-bridge"
Now, since we have
#2300.0cancel(g Na) * (cancel(1 mol Na))/(22.989 cancel(g Na)) * (cancel(1 mol H))/(cancel(1 mol Na)) * (1.0079 g H)/(cancel(1 mol H))#
#= color(blue)(100.84 "g" H(g))#
The theoretical yield of hydrogen gas is
Explanation:
We need to start with a balanced equation. This will give us the mole ratio between sodium metal and hydrogen gas. We will also need the molar masses of sodium and hydrogen gas in order to convert between moles and mass.
Balanced Equation
Mole ratio between
Molar masses of sodium and hydrogen gas
Convert
Convert moles of
Theoretical Yield of Hydrogen Gas
Convert moles of