Question

If 3.7115 moles of O2 are reacted, how many grams of CCl4 will be produced? a) 130.06 grams b) 91.84 grams c) 65.27 grams d) 47.35 grams

Answer 1

285.72 grams of
`CCl_4` will be produced. None of the given options is answer.

Step-by-step explanation:

The balanced chemical equation for the reaction between
`O_2` and
`CCl_4`is:

`2O_2` +
`CCl_4` →
`2CO_2` +
`2Cl_2`

According to this equation, 2 moles of
`O_2` react with 1 mole of
`CCl_4`. Therefore, 3.7115 moles of
`O_2` will react with:

3.7115 moles
`O_2` × (1 mole
`CCl_4`/ 2 moles
`O_2`) = 1.85575 moles
`CCl_4`

The molar mass of
`CCl_4` is 153.8 g/mol. Therefore, 1.85575 moles of
`CCl_4` will have a mass of:

1.85575 moles
`CCl_4` × 153.8 g/mol = 285.72 g

Therefore, the answer is 285.72 grams of
`CCl_4`.

None of the given options is answer.