Problem PageQuestion Liquid hexane CH3CH24CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseou

Question

nataly862011 [7]

3 years ago

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Problem PageQuestion Liquid hexane CH3CH24CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseou

s water H2O. Suppose 60. g of hexane is mixed with 74.5 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 3 significant digits.

Chemistry

2 answers:

const2013 [10] · Answer 1 · 2022-04-03T22:24:42+03:00

Answer:

$m_{H_2O}=30.9gH_2O$

Explanation:

Hello,

In this case, the combustion of hexane is given by:

$C_6H_{14}+\frac{19}{2} O_2\rightarrow 6CO_2+7H_2O$

The next step is to compute the reacting moles of hexane:

$n_{C_6H_{14}}=60gC_6H_{14}*\frac{1molC_6H_{14}}{86gC_6H_{14}} =0.698molC_6H_{14}$

Then, the moles of hexane consumed by 74.5 g of oxygen using the molar ratio in the chemical reaction (1:19/2):

$n_{C_6H_{14}}=74.5gO_2*\frac{1molO_2}{32gO_2} *\frac{1molC_6H_{14}}{19/2molO_2} =0.245molC_6H_{14}$

Therefore, as less moles of hexane are consumed by oxygen, it is in excess, so we compute the mass of water produced by the consumed 0.245 moles of hexane:

$m_{H_2O}=0.245molC_6H_{14}*\frac{7molH_2O}{1molC_6H_{14}}*\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=30.9gH_2O$