Question

Exactly 1.0 lb Hydrone, an alloy of sodium with lead, yields (at 0.0°C and 1.00 atm) 2.6 ft3 of hydrogen when it is treated with water. All the sodium reacts according to the following reaction: 2 Na 1 2 H O() 8n 2 NaOH(aq) 1 H (g) in alloy 2 2 and the lead does not react with water. Compute the per- centage by mass of sodium in the alloy.

Answer 1

Answer:

The percentage by mass of sodium in the alloy is 33.29%.

Explanation:

Volume of hydrogen gas = $V = 2.6 ft^3=73.6237 L$

$1 ft^3=28.3168 L$

Pressure of hydrogen gas = P = 1 atm

Temperature of the gas = T = 0.0°C =273.15 K

Moles of hydrogen gas = n

$PV=nRT$ (Ideal gas)

$n=\frac{PV}{RT}=\frac{1atm \times 73.6237 L}{0.0821 atm L/mol K\times 273.15 K}$

n = 3.2830 mole

Moles of hydrogen gas = 3.280 mole

$2 Na(s) +2H_2O(l)\rightarrow 2NaOH(aq)+ H_2 (g)$

According to reaction 1 mole of hydrogen is obtained from 2 moles of sodium.

Then 3.280 moles of hydrogen gas will be obtained from :

$\frac{2}{1}\times 3.280 mol=6.566 mol$

Mass of 6.566 moles of sodium =

6.566 mol × 23 g/mol = 151.02 g

Mass of hydrone = 1.0 lb = 453.592 g

The percentage by mass of sodium in the alloy:

$\frac{151.02 g}{ 453.592 g}\times 100=33.29\%$