Question

Hydrogen reacts w/ sodium to produce solid sodium hydride, NaH. A reaction mixture contains 6.75 g Na and 3.03 g H2.
A. Which reactant is limiting?
B. What is the theoretical yield of NaH from the above reaction mixture?
C. What is the percent yield if 4 g NaH is formed?


Laura, write the balanced equation:
2Na + H2 >>2NaH

It takes twice as many moles of sodium as hydrogen..
Figure the moles of sodium and hydrogen from the masses? If you have more than twice the moles of sodium as hydrogen, then the limiting reagent is hydrogen..think that out.
Now using the actual number of moles of the limiting reageant, what is the moles of product from the balanced equation?

I will be happy to critiqe your thinking.

Answer 1

2Na + H2 -> 2NaH

A. Trial 1:

n(Na) = mass/molar mass
n = 6.75/22.99
n = 0.294

n(Na):n(H2):n(NaH)
=2:1:2
=0.294:0.147:0.294

Trial 2:

n(H2) = mass/molar mass
n = 3.03/1.01
n = 3

n(Na):n(H2):n(NaH)
=2:1:2
=6:3:6

0.294<6
Therefore Na is the limiting reactant.

B. Theoretical yield:

(To reiterate)

n(Na) = mass/molar mass
n = 6.75/22.99
n = 0.294

n(Na):n(H2):n(NaH)
=2:1:2
=0.294:0.147:0.294

n(NaH) = 0.294

mass = n(molar mass)
m = 0.294(22.99 + 1.01)
m = 0.294(24)
m = 7.056g

Therefore, theoretical yield = 7.056g

C. %yield = actual yield/theoretical yield
%yield = 4/7.056
%yield = 0.567 = 56.7%