23 g of sodium shall react with methyl alcohol (CH3OH), to give

Question

23 g of sodium shall react with methyl alcohol (CH3OH), to give (A) one mole of oxygen (B) 1/2 mole of hydrogen (C) one mole of hydrogen (D) 2 g of hydrogen.

Tardigrade Admin · Accepted Answer

23 g of Na ≡ 1 mole of Na CH3OH + Na arrow CH3ONa + (1/2) H2 uparrow

Q. $23 g$ of sodium shall react with methyl alcohol $(C H_{3} O H)$ , to give

one mole of oxygen

1/2 mole of hydrogen

one mole of hydrogen

2 g of hydrogen.

$23 g$ of $N a \equiv 1$ mole of $N a$
$C H_{3} O H + N a \to C H_{3} ON a + \frac{1}{2} H_{2} ↑$

Q. 23g of sodium shall react with methyl alcohol (CH3​OH), to give