Answer:
[tex]mass of CHF_3 = 1/3 \times 9.47 \times 70 =220.97g[/tex]
Explanation:
[tex]CH_4(g)+3F_2(g)->3HF(g)+CHF_3[/tex]
As given in the question that methane([tex]CH_4[/tex]) is taken excess amount
mass of [tex]CHF_3[/tex] depend only on mass of fluorine
mass of [tex]F_2[/tex] =180 g
mole of [tex]F_2[/tex]=[tex]\frac{180}{19}[/tex] =9.47
3 mole of [tex]F_2[/tex] gives 1 mole of [tex]CHF_3[/tex]
so 1 mole will give [tex]\frac{1}{3}[/tex] mole of [tex]CHF_3[/tex]
therefore,
9.47 mole of [tex]F_2[/tex] will give [tex]\frac{1}{3} \times 9.47[/tex] mole of [tex]CHF_3[/tex]
[tex]mass of CHF_3 = 1/3 \times 9.47 \times 70 =220.97g[/tex]
