Question

g Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 0.96 g of methane is mixed with 6.37 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

Answer: 2.64 g of carbon dioxide that could be produced by the chemical reaction.

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of} CH_4=(0.96g)/(16g/mol)=0.06moles`

`\text{Moles of} O_2=(6.37)/(32)=0.20moles`

`CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)`

According to stoichiometry :

1 mole of
`CH_4` require = 2 moles of
`O_2`

Thus 0.06 moles of
`CH_4` will require=
`(2)/(1)* 0.06=0.12moles` of
`O_2`

Thus
`CH_4` is the limiting reagent as it limits the formation of product and
`O_2` is the excess reagent.

As 1 moles of
`CH_4` give = 1 mole of
`CO_2`

Thus 0.06 moles of
`CH_4` give =
`(1)/(1)* 0.06=0.06moles` of
`CO_2`

Mass of
`CO_2=moles* {\text {Molar mass}}=0.06moles* 44g/mol=2.64g`

Thus 2.64 g of carbon dioxide that could be produced by the chemical reaction.