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blagie [28]
3 years ago
14

Calculate the mass of water produced when 3.03 g of butane reacts with excess oxygen.

Chemistry
1 answer:
algol133 years ago
6 0

The mass of water produced is 4.70 g.  

<em>M</em>_r:   58.12     32.00  

         2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

<em>m</em>/g:  3.03

<em>Moles of C₄H₁₀</em> = 3.03 g C₄H₁₀ × (1 mol C₄H₁₀/58.12 g C₄H₁₀) = 0.05 213 mol C₄H₁₀

<em>Moles of H₂O</em> = 0.052 13 mol C₄H₁₀ × (10 mol H₂O/2 mol C₄H₁₀) = 0.2607 mol H₂O

Mass of H₂O = 0.2607 mol H₂O × (18.02 g H₂O/1 mol H₂O) = 4.70 g H₂O

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