Question

question 3 (2 points)
how many grams of o₂ (mm = 32.00 g/mol) are consumed during the complete combustion of 2.62 grams of c₃h₈ (mm = 44.11 g/mol)? equation is given below.
report answer to 3 significant figures in grams, but do not include the unit in your response.

c₃h₈ (g) + 5 o₂ (g) → 3 co₂ (g) + 4 h₂o (g)

Explanation:

Step1: Calculate moles of C₃H₈

Use the formula $n=\frac{m}{M}$, where $n$ is moles, $m$ is mass and $M$ is molar - mass.
$n_{C_3H_8}=\frac{2.62\ g}{44.11\ g/mol}=0.0594\ mol$

Step2: Determine moles of O₂ based on stoichiometry

From the balanced equation $C_3H_8(g)+5O_2(g)\to3CO_2(g) + 4H_2O(g)$, the mole - ratio of $C_3H_8$ to $O_2$ is $1:5$.
$n_{O_2}=5\times n_{C_3H_8}=5\times0.0594\ mol = 0.297\ mol$

Step3: Calculate mass of O₂

Use the formula $m = n\times M$, where $n$ is moles and $M$ is molar - mass.
$m_{O_2}=n_{O_2}\times M_{O_2}=0.297\ mol\times32.00\ g/mol = 9.50$

Answer:

9.50