Question

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?

Answer 1

The sample contains 50.5 % Na by mass.

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

`PV = nRT`

`n = (PV)/(RT) = (100.0"kPa" × 1.164"L")/(8.314"kPa·L·K⁻¹mol⁻¹" × 300.0"K")` = 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 × `(1"mol Na")/(22.99"g Na") × (1"mol H₂")/(2"mol Na")` = 0.0217 49`x` mol H₂

moles of H₂ from Ca = (2.00 - `x`) g Ca × `(1"mol Ca")/(40.08"g Na") × (2"mol H₂")/(2"mol Ca")` =

(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 = (0.003 23)/(0.003 201)` = 1.01

Mass of Na = 1.01 g

3. Calculate the % of Na by mass.

% Na = `(1.01"g")/(2.00"g")` × 100 % = 50.5 %