Question

I NEED HELP PLEASE, THANKS! :)

Combustion reactions are a notable source of carbon dioxide in the environment. Using the following balanced equation, how many grams of carbon dioxide are formed when 100.00 g of propane (C3H8) is burned? Express your answer to the correct number of significant figures.

Answer 1

Number of moles of propane:

=Mass in grams ÷ Relative molecular Mass

= 100/((12*3) + (1*8))

= 100 ÷ 44

= 2.2727

Mole ratio propane:carbon (IV) oxide = 1:3(from the equation)

Number of moles of CO2 = 3*2.2727 = 6.8181

Mass in grams = Relative molecular Mass * Number of moles

= 44 * 6.8181

= 299.9964 grams

Step-by-step explanation:

Answer 2

`\large \boxed{\text{299.4 g}}`

Step-by-step explanation:

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 44.10 44.01

C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O

m/g: 100.00

To solve a stoichiometry problem, you must

1. Use the molar mass to convert mass of C₃H₈ to moles of C₃H₈
2. Use the molar ratio to convert moles of C₃H₈ to moles of CO₂
3. Use the molar mass to convert moles of CO₂ to mass of CO₂

1. Moles of C₃H₈

`\text{Moles of C$_(3)$H}_(8) = \text{100.00 g C$_(3)$H}_(8) * \frac{\text{1 mol C$_(3)$H}_(8)}{\text{44.10 g C$_(3)$H}_(8)} = \text{2.268 mol C$_(3)$H}_(8)`

2. Moles of CO₂

The molar ratio is 3 mol CO₂:1 mol C₃H₈

`\text{Moles of CO}_(2) = \text{2.268 mol C$_(3)$H}_(8) * \frac{\text{3 mol CO}_(2)}{\text{1 mol C$_(3)$H}_(8)} = \text{6.803 mol CO}_(2)`

3. Mass of CO₂

`\text{Mass of CO}_(2) = \text{6.803 mol CO}_(2) * \frac{\text{44.01 g CO}_(2)}{\text{1 mol CO}_(2)} = \textbf{299.4 g CO}_(2)\\\text{The mass of CO$_(2)$ required is $\large \boxed{\textbf{299.4 g}}$}`