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forsale [732]
3 years ago
6

Calculate the mass of water produced when 9.57 g of butane reacts with excess oxygen.

Chemistry
1 answer:
irinina [24]3 years ago
4 0

Answer:

14.9 g

Explanation:

Step 1: Write the balanced equation

C₄H₁₀ + 6.5 O₂ ⇒ 4 CO₂ + 5 H₂O

Step 2: Calculate the moles corresponding to 9.57 g of C₄H₁₀

The molar mass of C₄H₁₀ is 58.12 g/mol.

9.57 g × 1 mol/58.12 g = 0.165 mol

Step 3: Calculate the moles of H₂O produced from 0.165 moles of C₄H₁₀

0.165 mol C₄H₁₀ × 5 mol H₂O/1 mol C₄H₁₀ = 0.825 mol H₂O

Step 4: Calculate the mass corresponding to 0.825 mol of H₂O

The molar mass of H₂O is 18.02 g/mol.

0.825 mol × 18.02 g/mol = 14.9 g

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<u>Answer:</u> The mass of decane produced is 1.743\times 10^2g

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To calculate the number of moles, we use the equation:  

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As, decene is present in excess. So, it is considered as an excess reagent.

Thus, hydrogen gas is a limiting reagent because it limits the formation of products.

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So, 1.225 moles of hydrogen gas will produce = \frac{1}{1}\times 1.225=1.225mol of decane

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1.225mol=\frac{\text{Mass of decane}}{142.30g/mol}\\\\\text{Mass of carbon dioxide}=(1.225mol\times 142.30g/mol)=174.3g=1.743\times 10^2g

Hence, the mass of decane produced is 1.743\times 10^2g

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