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NNADVOKAT [17]
3 years ago
8

Calculate the mass of water produced when 2.06 g of butane reacts with excess oxygen.

Chemistry
1 answer:
dem82 [27]3 years ago
8 0
2C4H10 + 13O2 = 8CO2 + 10H2O

1. (2.06g C4H10)/(58.12 g/mol C4H10) = 0.035mol C4H10

2. (0.035molC4H10)(10 mol H2O/2mol C4H10) = 0.177mol H2O

3. (0.177mol H2O)(18.01g/mol H2O) = 3.19g H2O
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<h3><u>Answer;</u></h3>

= 930.23 mL

<h3><u>Explanation</u>;</h3>

Using the combined gas law;

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Where; P1 = 600 kPa, V1 = 800 mL, and T1 = -25 +273 = 258 K, and

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Thus;

V2 = P1V1T2/T1P2

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