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FromTheMoon [43]
3 years ago
14

An automobile gasoline tank holds 42 kg of gasoline. When the gasoline burns, 168 kg of oxygen are consumed and carbon dioxide a

nd water are produced. What total combined mass of carbon dioxide and water is produced?
Chemistry
1 answer:
Georgia [21]3 years ago
5 0

Answer:

210 kg

Explanation:

The chemical reaction will look like this:

gasoline + O₂ (oxygen) → CO₂ (carbon dioxide) + H₂O (water) + heat

Now in chemical reactions the quantity of the reactants (gasoline + oxygen) is equal to the quantity of products (carbon dioxide + water), so from this principle we may determine:

combined mass of carbon dioxide and water produced = mass of gasoline + mass of oxygen

combined mass of carbon dioxide and water produced = 42 + 168  = 210 kg

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4.083 * 10^20 atoms.

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6.022 * 10^23  * 0.021 / 30.974

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Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.7 g of ethane is m
Bond [772]

Answer:

m_{H_2O}=4.86gH_2O

Explanation:

Hello,

In this case, the described chemical reaction is:

C_2H_6+\frac{7}{2} O_2\rightarrow 2CO_2+3H_2O

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

n_{C_2H_6}=2.7gC_2H_6*\frac{1molC_2H_6}{30gC_2H_6} =0.09molC_2H_6

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

n_{C_2H_6}^{consumed\ by \ O_2}=13.0gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6}{\frac{7}{2} molO_2}=0.116molC_2H_6

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

m_{H_2O}=0.09molC_2H_6*\frac{3molH_2O}{1molC_2H_6} *\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=4.86gH_2O

Best regards.

3 0
3 years ago
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