Question

Given the reaction represented by the balanced equation:

CH4(g)+3F2(g)->3HF(g)+CHF3 calculate the number of g of CHF3 that could be produced by mixing 180 g F2 with excess CH4.

Answer 1

`mass of CHF_3 = 1/3 * 9.47 * 70 =220.97g`

Step-by-step 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`=
`(180)/(19)` =9.47

3 mole of
`F_2` gives 1 mole of
`CHF_3`

so 1 mole will give
`(1)/(3)` mole of
`CHF_3`

therefore,

9.47 mole of
`F_2` will give
`(1)/(3) * 9.47` mole of
`CHF_3`

`mass of CHF_3 = 1/3 * 9.47 * 70 =220.97g`