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forsale [732]
1 year ago
12

Equation is balanced already

Chemistry
1 answer:
mojhsa [17]1 year ago
6 0

Answer:

1.) 13 g C₄H₁₀

2.) 41 g CO₂

Explanation:

To find the mass of propane (C₄H₁₀) and carbon dioxide (CO₂), you need to (1) convert mass O₂ to moles O₂ (via molar mass), then (2) convert moles O₂ to moles C₄H₁₀/CO₂ (via mole-to-mole ratio from equation coefficients), and then (3) convert moles C₄H₁₀/CO₂ to mass C₄H₁₀/CO₂ (via molar mass). It is important to arrange the ratios in a way that allows for the cancellation of units. The final answers should have 2 sig figs to match the sig figs of the given value.

Molar Mass (C₄H₁₀): 4(12.011 g/mol) + 10(1.008 g/mol)

Molar Mass (C₄H₁₀): 58.124 g/mol

Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)

Molar Mass (CO₂): 44.007 g/mol

Molar Mass (O₂): 2(15.998 g/mol)

Molar Mass (O₂): 31.996 g/mol

2 C₄H₁₀ + 13 O₂ ----> 8 CO₂ + 10 H₂O

 48 g O₂             1 mole             2 moles C₄H₁₀            58.124 g
---------------  x  -----------------  x  -------------------------- x  ------------------  =  
                         31.996 g              13 moles O₂               1 mole

=  13 g C₄H₁₀

 48 g O₂             1 mole               8 moles CO₂            44.007 g
---------------  x  -----------------  x  -------------------------- x  ------------------  =  
                         31.996 g              13 moles O₂               1 mole

=  41 g CO₂

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Combustion of 9.511 grams of c4h10 will yield ____ grams of CO2
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\boxed{28.81}

Explanation:

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           2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

m/g:     9.511

1. Moles of C₄H₁₀

\text{Moles of C$_{4}$H$_{10} $} = \text{ 9.511 g C$_{4}$H$_{10} $} \times \dfrac{\text{1 mol C$_{4}$H$_{10} $}}{\text{ 58.12 g C$_{4}$H$_{10} $}} = \text{0.1636 mol C$_{4}$H$_{10}$}

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\text{Moles of CO}_{2} =\text{0.1636 mol C$_{4}$H$_{10} $} \times \dfrac{\text{8 mol CO}_{2}}{\text{2 mol C$_{4}$H$_{10}$}} = \text{0.6546 mol CO}_{2}

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\text{Mass of CO}_{2} = \text{0.6546 mol CO}_{2} \times \dfrac{\text{44.01 g CO}_{2}}{\text{1 mol CO}_{2}} = \textbf{28.81 g CO}_{2}\\\\\text{The combustion will form $\boxed{\textbf{28.81 g CO}_{2}}$}

8 0
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