The mass of water produced is 792 grams by the combustion of 568 grams of decane.
Given:
Combustion of 568 grams of decane with 2979 grams of oxygen.
[tex]2 C_{10}H_{22 }+ 31 O_2 \rightarrow 20 CO_2 + 22 H_2O[/tex]
To find:
The mass of water produced by combustion of 568 grams of decane.
Solution:
Mass of decane = 568 g
Moles of decane :
= [tex]\frac{568 g}{142 g/mol}=4 mol[/tex]
Mass of oxygen gas = 2976 g
Moles of oxygen gas:
= [tex]\frac{2976 g}{32 g/mol}=93 mol[/tex]
[tex]2 C_{10}H_{22 }+ 31 O_2 \rightarrow 20 CO_2 + 22 H_2O[/tex]
According to reaction, 2 moles of decane reacts with 31 moles of oxygen, then 4 moles of decane will react with:
[tex]=\frac{31}{2}\times 4mol=62\text{ mol of}O_2[/tex]
But according to the question, we have 93.0 moles of oxygen gas which is more than 62 moles of oxygen gas.
So, this means that oxygen gas is present in an excessive amount. Which simply means:
- Oxygen gas is an excessive reagent.
- Decane is a limiting reagent.
- Decane being limiting reagent will be responsible for the amount of water produced after the reaction.
According to reaction, 22 moles of water is produced from 2 moles of decane, then 4 moles of decane will produce:
[tex]=\frac{22}{2}\times 4mol=44\text{mol of }H_2O[/tex]
Mass of 44 moles of water ;
[tex]=44mol\times 18g/mol=792g[/tex]
792 grams of water is produced by the combustion of 568 grams of decane.
Learn more about limiting reagent and excessive reagent here:
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