# When 2.00 g mixture of Na and Ca reat with water, 1.164 L hydrogen was produced at 300.0 K and 100.0 kPa. What is the percentage of Na in the sample?

Jul 10, 2014

The sample contains 50.5 % Na by mass.

1. Use the Ideal Gas Law to calculate the moles of hydrogen.

$P V = n R T$

$n = \frac{P V}{R T} = \left(100.0 \text{kPa" × 1.164"L")/(8.314"kPa·L·K⁻¹mol⁻¹" × 300.0"K}\right)$ = 0.0466 68 mol H₂

(4 significant figures + 1 guard digit)

2. Calculate the moles of Na and Ca (This is the tough part).

The balanced equations are

2Na + 2H₂O → 2NaOH + H₂
2Ca + 2H₂O → Ca(OH)₂ + 2H₂

Let mass of Na = $x$ g. Then mass of Ca = (2.00 - $x$) g

moles of H₂ = moles of H₂ from Na + moles of H₂ from Ca

moles of H₂ from Na = $x$ g Na × $\left(1 \text{mol Na")/(22.99"g Na") × (1"mol H₂")/(2"mol Na}\right)$ = 0.0217 49$x$ mol H₂

moles of H₂ from Ca = (2.00 - $x$) g Ca × $\left(1 \text{mol Ca")/(40.08"g Na") × (2"mol H₂")/(2"mol Ca}\right)$ =

(0.049 90 – 0.024 950$x$) mol H₂

moles of H₂ from Na + moles of H₂ from Ca = total moles of H₂

0.0217 49$x$ mol H₂ + (0.049 90 – 0.0249 50$x$) mol H₂ = 0.0466 68 mol H₂

0.0217 49$x$ + 0.049 90 - 0.0249 50$x$ = 0.0466 68

0.003 201$x$ = 0.003 23

$x = \frac{0.003 23}{0.003 201}$ = 1.01

Mass of Na = 1.01 g

3. Calculate the % of Na by mass.

% Na = $\left(1.01 \text{g")/(2.00"g}\right)$ × 100 % = 50.5 %