Question

Given the reaction represented by the balanced equation:

Answer 1

Answer:

$mass of CHF_3 = 1/3 \times 9.47 \times 70 =220.97g$

Explanation:

$CH_4(g)+3F_2(g)->3HF(g)+CHF_3$

As given in the question that methane($CH_4$) is taken excess amount

mass of $CHF_3$ depend only on mass of fluorine

mass of $F_2$ =180 g

mole of $F_2$=$\frac{180}{19}$ =9.47

3 mole of $F_2$ gives 1 mole of $CHF_3$

so 1 mole will give $\frac{1}{3}$ mole of $CHF_3$

therefore,

9.47 mole of $F_2$ will give $\frac{1}{3} \times 9.47$ mole of $CHF_3$

$mass of CHF_3 = 1/3 \times 9.47 \times 70 =220.97g$