# 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.

## 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%

## Your thinking is correct. Let's calculate the moles of sodium and hydrogen in the reaction mixture.

The molar mass of sodium (Na) is 22.99 g/mol and the molar mass of hydrogen (H2) is 2.02 g/mol.

A. Moles of sodium (Na):
6.75 g Na * (1 mol Na / 22.99 g Na) = 0.293 mol Na

B. Moles of hydrogen (H2):
3.03 g H2 * (1 mol H2 / 2.02 g H2) = 1.50 mol H2

From the balanced equation, we can see that it takes twice as many moles of sodium as hydrogen to react. Therefore, if the moles of sodium are more than twice the moles of hydrogen, hydrogen is the limiting reactant. Let's check:

2 * 1.50 mol H2 = 3.00 mol H2
0.293 mol Na < 3.00 mol H2

Since the moles of sodium are less than twice the moles of hydrogen, sodium is the limiting reactant.

C. Moles of sodium hydride (NaH) from the balanced equation:
2 mol NaH = 2 mol Na (1 mol NaH / 2 mol Na) = 0.293 mol NaH

Now we can calculate the theoretical yield of NaH.

The molar mass of sodium hydride (NaH) is 23.00 g/mol.

B. Theoretical yield of NaH:
0.293 mol NaH * (23.00 g NaH / 1 mol NaH) = 6.74 g NaH

C. Percent yield if 4 g NaH is formed:
Percent yield = (Actual yield / Theoretical yield) * 100
Percent yield = (4 g NaH / 6.74 g NaH) * 100 = 59.27%

## To determine the limiting reactant, you need to compare the moles of sodium and hydrogen from the given masses.

1. Moles of sodium (Na):
The molar mass of sodium (Na) is 22.99 g/mol.
Moles of Na = Mass of Na (in grams) / Molar mass of Na
Moles of Na = 6.75 g / 22.99 g/mol = 0.293 mol

2. Moles of hydrogen (H2):
The molar mass of hydrogen (H2) is 2.02 g/mol.
Moles of H2 = Mass of H2 (in grams) / Molar mass of H2
Moles of H2 = 3.03 g / 2.02 g/mol = 1.50 mol

Since the moles of Na (0.293 mol) is less than twice the moles of H2 (1.50 mol), hydrogen is the limiting reactant.

Now, using the actual number of moles of the limiting reactant (hydrogen), you can determine the moles of product (NaH) from the balanced equation.

The balanced equation is: 2Na + H2 â†’ 2NaH

From the equation, you can see that 1 mole of H2 reacts to form 2 moles of NaH.
So, the moles of NaH formed can be calculated as follows:

Moles of NaH = Moles of H2 Ã— (2 moles of NaH / 1 mole of H2)
Moles of NaH = 1.50 mol Ã— (2 mol NaH / 1 mol H2) = 3.00 mol NaH

Therefore, the theoretical yield of NaH from the given reaction mixture is 3.00 moles.

Finally, to calculate the percent yield, you need to compare the actual yield (given as 4 g NaH) with the theoretical yield (calculated above).

1. Moles of actual yield of NaH:
The molar mass of NaH is 23.00 g/mol.
Moles of NaH = Mass of NaH (in grams) / Molar mass of NaH
Moles of NaH = 4 g / 23.00 g/mol = 0.174 mol

2. Percent yield:
Percent yield = (Actual yield / Theoretical yield) Ã— 100
Percent yield = (0.174 mol / 3.00 mol) Ã— 100 = 5.8%

Therefore, the percent yield of NaH in this reaction is 5.8%.