Question

Read the given equation. 2Na + 2H2O → 2NaOH + H2

During a laboratory experiment, a certain quantity of sodium metal reacted with water to produce sodium hydroxide and hydrogen gas. What was the initial quantity of sodium metal used if 6.30 liters of H2 gas were produced at STP?
A.) 10.3 grams
B.) 12.9 grams
C.) 14.7 grams
D.) 15.2 grams

Answer 1

B) 12.9 grams.

Step-by-step explanation:

How many moles of molecules in that 6.30 L of H₂?

The volume of one mole of an ideal gas at STP (0 °C, 1 atm) is 22.4 liters.

(The volume of that one mole of gas at STP will be 22.7 liters if STP is defined as 0 °C and 10⁵ Pa).

`\displaystyle n(\text{H}_2) = \frac{6.30\;\text{L}}{22.4\;\text{L}\cdot\text{mol}^(-1)} = 0.28125\;\text{mol}`.

How many moles of Na will be needed?

The coefficient in front of Na in the equation is twice the coefficient in front of H₂. It takes two moles of Na to produce one mole of H₂.

`n(\text{Na}) = 2\;n(\text{H}_2) = 2* 0.28125= 0.5625\;\text{mol}`.

What's the mass of that many Na atoms?

Refer to a modern periodic table. The molar mass of ₁₁Na is 22.990. The mass of one mole of Na atoms is 22.990 gram. The mass of 0.5625 moles of Na atoms will be

`m = n\cdot M = 0.5625 * 22.990 = 12.9\;\text{g}`.

(2 sig. fig. as in the volume of the H₂ gas.)