What is the quantity of hydrogen gas liberated when `46 g` sodium reacts with excess ethanol ?
(Given atomic mass of `Na = 23`)

Options

A. `2.4 xx 10^(-3)` kg
B. `2.0 xx 10^(-3)` kg
C. `4.0 xx 10^(-3)` kg
D. `2.4 xx 10^(-2)` kg

Answer

B

Explanation

`2C_(2)H_(5)OH + 2Na to underset("sod. Ethoxide") (2C_(2)H_(5)Ona + H_(2)) uarr`
46 of ef Na means 2 moles of Na liberates one mole of `H_(2)` i.e., 1g molar mass i.e., i.e. `2 xx 10^(3)` kg.