Question

Gaseous methane (CH4) reacts with gaseous oxygen (O2) gas to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). What is the theoretical yield of carbon dioxide formed from the reaction of 1.28 g of methane and 10.1 g of oxygen gas? Be sure your answer has the correct number of significant digits in it.

Answer 1

Theoretical yield of
`CO_(2)` is 3.51 g.

Step-by-step explanation:

Balanced equation:
`CH_(4)+2O_(2)\rightarrow CO_(2)+2H_(2)O`

Molar mass of
`CH_(4)` = 16.04 g/mol

Molar mass of
`O_(2)` = 32.00 g/mol

So, 1.28 g of
`CH_(4)` =
`(1.28)/(16.04)mol` of
`CH_(4)` = 0.0798 mol of
`CH_(4)`

10.1 g of
`O_(2)` =
`(10.1)/(32.00)mol` of
`O_(2)` = 0.316 mol of
`O_(2)`

According to balanced equation-

1 mol of
`CH_(4)` produces 1 mol of
`CO_(2)`

So, 0.0798 mol of
`CH_(4)` produce 0.0798 mol of
`CO_(2)`

2 moles of
`O_(2)` produce 1 mol of
`CO_(2)`

So, 0.316 mol of
`O_(2)` produce 0.158 mol of
`CO_(2)`

As least number of moles of
`CO_(2)` are produced from
`CH_(4)` therefore
`CH_(4)` is the limiting reagent.

So, theoretical yield of
`CO_(2)` = 0.0798 mol

Molar mass of
`CO_(2)` = 44.01 g/mol

So, theoretical yield of
`CO_(2)` =
`(44.01* 0.0798)g=3.51g`