Question

chemistry 20 stoichiometry quiz name: ebtesam numeric response use the following information to answer the next question. the combustion of butane gas is represented by this equation: _c₄h₁₀(g) + _o₂(g) → _co₂(g) + _h₂o(g) numeric response 3. how many grams of co₂(g) are produced when 1.00 g of c₄h₁₀(g) react with the excess oxygen? (record your answer in the numerical - response section below.) your answer:

Explanation:

Step1: Balance the chemical equation

$2C_{4}H_{10}(g)+13O_{2}(g)
ightarrow8CO_{2}(g)+10H_{2}O(g)$

Step2: Calculate the molar mass of $C_{4}H_{10}$ and $CO_{2}$

The molar mass of $C_{4}H_{10}$: $M_{C_{4}H_{10}}=(4\times12.01 + 10\times1.01)\ g/mol=58.14\ g/mol$. The molar mass of $CO_{2}$: $M_{CO_{2}}=(12.01+2\times16.00)\ g/mol = 44.01\ g/mol$.

Step3: Calculate the moles of $C_{4}H_{10}$

$n_{C_{4}H_{10}}=\frac{m_{C_{4}H_{10}}}{M_{C_{4}H_{10}}}=\frac{1.00\ g}{58.14\ g/mol}\approx0.0172\ mol$

Step4: Determine the moles of $CO_{2}$ produced

From the balanced - equation, the mole ratio of $C_{4}H_{10}$ to $CO_{2}$ is $2:8 = 1:4$. So, $n_{CO_{2}} = 4\times n_{C_{4}H_{10}}=4\times0.0172\ mol = 0.0688\ mol$

Step5: Calculate the mass of $CO_{2}$ produced

$m_{CO_{2}}=n_{CO_{2}}\times M_{CO_{2}}=0.0688\ mol\times44.01\ g/mol\approx3.03\ g$

Answer:

$3.03$