Question

Question #c373d

Answer 1

Here is the equation:
`C_2H_4 + 3O_2 rarr 2CO_2 + 2H_2O`

Other solutions follow...

Explanation:

In the equation, we see that each one mole of `C_2H_4` results in two moles of `CO_2`. THis establishes the ratio between these compounds. So, 0.02 mole of `C_2H_4` must yield 0.04 mol of `CO_2`

Since the molar mass of `CO_2` is 44.0 g, the mass of 0.04 mol is
`0.04 xx 44.0 = 1.76 g`

The volume of the gas at r.t.p. (can I assume this is room temperature of 25°C?) is found by multiplying by the molar volume of a gas at 25°C, namely 24.5 L

`0.04 "mol" xx 24.5 L/"mol" = 0.98 L`