Question

Be sure to answer all parts. Propane (C3H8) is a minor component of natural gas and is used in domestic cooking and heating. (a) Balance the following equation representing the combustion of propane in air. Include states of matter in your answer. C3H8(g) + O2(g) → CO2(g) + H2O(g) (b) How many grams of carbon dioxide can be produced by burning 8.11 moles of propane? Assume that oxygen is the excess reactant in this reaction. × 10 g Enter your answer in scientific notation.

Answer 1

Answer:

For a: The balanced chemical equation is given below.

For b: The mass of carbon dioxide produced will be $1.07\times 10^3g$

Explanation:

• For a:

Every balanced chemical equation follows law of conservation of mass.

This law states that mass can neither be created nor can be destroyed but it can only be transformed from one form to another form.

This law also states that the total number of individual atoms on the reactant side must be equal to the total number of individual atoms on the product side.

For the given reaction, the balance chemical equation follows:

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)$

All the substances are present in gaseous state.

• For b:

By Stoichiometry of the reaction:

1 mole of propane gas produces 3 moles of carbon dioxide gas.

So, 8.11 moles of propane gas will produce = $\frac{3}{1}\times 8.11=24.33mol$ of carbon dioxide gas.

Now, calculating the mass of carbon dioxide using equation:

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

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 24.33 mol

Putting values in above equation, we get:

$24.33mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=1070.52g$

Hence, the amount of $CO_2$ produced in the given reaction and expressed in scientific notation is $1.07\times 10^3g$