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Bas_tet [7]
3 years ago
11

Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose

58. g of octane is mixed with 133. g of oxygen. Calculate the minimum mass of octane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
Olin [163]3 years ago
6 0

2C6H14 + 19O2 ===> 12CO2 + 14H2O ... balanced equation

moled hexane present = 2.6 g x 1 mole/130 g = 0.02 moles

moles O2 present = 5.29 g x 1 mole/32 g = 0.165 moles

Which reactant is limiting? Hexane = 0.02/2 = 0.01; O2 = 0.165/19 = 0.0087

Thus O2 is limiting...

moles of H2O that can be produced =0.165 moles O2 x 14 H2O/19 CO2 = 0.122 moles H2O

Mass H2O = 0.122 moles x 18 gm/mole = 2.20 g (to 3 sig. figs.)

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3 years ago
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How many moles are in 5.5 x 10-23 molecules of H2O
masha68 [24]

Answer:

0.914moles

Explanation:

The number of moles in a substance can be got by dividing the number of atoms/molecules/particles by Avagadro's constant (6.02 × 10^23).

That is;

number of moles (n) = number of atom (nA) ÷ 6.02 × 10^23

According to this question, there are 5.5 x 10-23 molecules of H2O

n = 5.5 x 10^23 ÷ 6.02 × 10^23

n = 0.914 × 10^(23-23)

n = 0.914 × 10^0

n = 0.914 × 1

n = 0.914moles

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Answer:

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8 0
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Helga [31]

Answer:

A. 1 = 3 - 2

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So if we look at the letter options we can see that A is the best choice.

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