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

Answer: (B)

$2.0 \times 10^{-3} kg$

Answer 2

$2 Na +2 C _{2} H _{5}- OH \rightarrow 2 C _{2} H _{5}- ONa + H _{2}$
Moles of $46\, g\, Na =\frac{46\, g }{23\, g / mol }=2\, mol$
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 $2.0 \times 10^{-3} kg$