Question

If $60$ grams of $NO$ is reacted with sufficient amount of $O_2$ to form $N{O}_2$ which is removed during the reaction, how many grams of $N{O}_2$ can be produced?

(Molar masses : $NO$ $=30g/mol$, $N{O}_2$ $=46g/mol$)

• A. $46$ g
• B. $60$ g
• C. $92$ g
• D. $120$ g
• E. $180$ g

Answer 1

The correct option is C $92$ g

$NO+\frac{1}{2}{O}_2\to N{O}_2$

Molar masses: $NO$ $=$ $30g/mol,$ NO${}_2$ $=$ $46g/mol$.

$1$ mole of $NO$ gives $1$ mole of $N{O}_2$

$30g$ of $NO$ gives $46g$ of $N{O}_2$.

Thus $60g$ of $NO$ will give $\frac{60}{30}\times 46=92$ g of $N{O}_2$.