Question

What is the quantity of hydrogen gas liberated when 46 g sodium reacts with excess ethanol ?

[Given At. mass of Na = 23]

• A. $2.4\times {10}^{-3}kg$
• B. $2.0\times {10}^{-3}kg$
• C. $4.0\times {10}^{-3}kg$
• D. $2.4\times {10}^{-2}kg$

Answer 1

The correct option is B $2.0\times {10}^{-3}kg$

$2Na+2C_2{H}_5-OH\to 2C_2{H}_5-ONa+H_2$

Moles of 46 g Na $=\frac{46g}{23g/mol}=2mol$

$2$ moles of Na will give $1$ mole of hydrogen.

The molecular weight of hydrogen is $2$ g/mol.

Thus the quantity of hydrogen gas liberated is 2.0 g or $2.0\times {10}^{-3}kg$