Question

As with any combustion reaction, the products of combusting a hydrocarbon fuel (CxHy) with oxygen (O2) are carbon dioxide (CO2) and water (H2O).

A mass of 15.51 g for an unknown fuel was combusted in a reaction vessel containing an unknown amount of oxygen. At the end of the reaction, there still remained 15.46 g of the fuel as well as 0.0817 g of water and 0.1497 g of carbon dioxide. The oxygen was completely consumed during the reaction.

How many molecules of oxygen gas were initially present in the reaction vessel?

Answer 1

Step-by-step explanation:

Ambitious

The basic law of mass conservation states that, the total mass of products of a chemical reaction should be equal to the total mass of reactants in the same reaction.

Applying this law to the above problem, we find that:

Total mass of reactants = 46 + 96 = 142 grams

This means that the total mass of products (water + carbon dioxide) should be also 142 grams

Since 52 grams of water is produced, therefore,

mass of carbon dioxide = 142 - 52 = 90 grams

Answer 2

Answer: The number of oxygen gas molecules initially present are
`3.409* 10^(21)`

Step-by-step explanation:

The balanced chemical reaction for the combustion of given hydrocarbon follows:

`C_xH_y+(x+(y)/(4))O_2\rightarrow xCO_2+(y)/(2)H_2O`

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

• For carbon dioxide:

Given mass of carbon dioxide = 0.1497 g

Molar mass of carbon dioxide = 44 g/mol

Putting values in above equation, we get:

`\text{Moles of carbon dioxide}=(0.1497g)/(44g/mol)=3.403* 10^(-3)mol`

Thus,
`x=3.402* 10^(-3)mol`

• For water:

Given mass of water = 0.0817 g

Molar mass of water = 18 g/mol

Putting values in above equation, we get:

`\text{Moles of water}=(0.0817g)/(18g/mol)=4.54* 10^(-3)mol`

Thus,
`(y)/(2)=4.54* 10^(-3)mol`

To calculate the number of moles of oxygen, we solve the equation for its coefficient by putting values of 'x' and 'y', we get:

`\Rightarrow (3.402* 10^(-3))+((4.54* 10^(-3))/(2))=5.662* 10^(-3)mol`

According to mole concept:

1 mole of a compound contains
`6.022* 10^(23)` number of molecules.

So,
`5.662* 10^(-3)` moles of oxygen will contain =
`5.662* 10^(-3)* 6.022* 10^(23)=3.409* 10^(21)` molecules.

Hence, the number of oxygen gas molecules initially present are
`3.409* 10^(21)`