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Answer :
Final answer:
The maximum mass of water that could be produced from the reaction of 20.0 g of butane and 37.9 g of oxygen is 16.43 g. Oxygen is the limiting reagent in this reaction. The result is based on the stoichiometry of the balanced equation and considering the significant digits of the initial data.
Explanation:
When considering the reaction between gaseous butane (CH3CH2CH2CH3) and gaseous oxygen (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O), we identify the limiting reagent to determine the maximum mass of water produced. The balanced chemical equation for the combustion of butane is:
C4H10(g) + 13/2 O2(g) → 4 CO2(g) + 5 H2O(g)
To find the maximum mass of water that can be formed, we first calculate the moles of butane and oxygen. Using the molar masses (butane: 58.12 g/mol, oxygen: 32.00 g/mol), we determine:
- For 20.0 g butane: 20.0 g / 58.12 g/mol = 0.344 moles of butane
- For 37.9 g oxygen: 37.9 g / 32.00 g/mol = 1.184 moles of oxygen
Using the stoichiometry of the balanced equation, 1 mole of butane reacts with 13/2 moles of oxygen. Thus, 0.344 moles of butane would require 0.344 × (13/2) = 2.236 moles of oxygen, which exceeds the available oxygen. Hence, oxygen is the limiting reagent.
Now, since 13/2 moles of oxygen produce 5 moles of water, we calculate the moles of water produced by 1.184 moles of oxygen:
1.184 moles O2 × (5 moles H2O / (13/2) moles O2) = 0.912 moles H2O
Converting moles of water to mass:
0.912 moles H2O × 18.02 g/mol = 16.43 g of H2O
Therefore, the maximum mass of water that could be produced is 16.43 g, with the amount of significant digits reflecting the precision of the given data.
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