Let's break down each part of your question step by step.
1. Stoichiometry of Butane Combustion
The combustion reaction of butane \((\text{C}4\text{H}{10})\) can be written as follows:
\[ \text{C}4\text{H}{10} + O_2 \rightarrow CO_2 + H_2O \]
The balanced equation is:
\[ 2 \text{C}4\text{H}{10} + 13 O_2 \rightarrow 8 CO_2 + 10 H_2O \]
From the balanced equation:
- 2 moles of butane react with 13 moles of oxygen.
Now, to find the mass of oxygen needed per kilogram of butane:
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Calculate the molar mass of butane \((\text{C}4\text{H}{10})\): \[ \text{Molar mass of C}4\text{H}{10} = (4 \times 12.01) + (10 \times 1.008) = 58.12 , \text{g/mol} \]
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The total moles of oxygen required for 1 mole of butane: For 2 moles of butane, 13 moles of oxygen are needed: \[ \text{For } 1 \text{ mole of butane: } \frac{13}{2} = 6.5 \text{ moles of } O_2 \]
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Find the molar mass of oxygen: \[ \text{Molar mass of } O_2 = 2 \times 16 = 32 , \text{g/mol} \]
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Calculate the mass of oxygen required for combustion of 1 kg of butane:
- First, find the moles in 1 kg of butane: \[ \text{Moles of butane in 1000g} = \frac{1000 , \text{g}}{58.12 , \text{g/mol}} \approx 17.19 , \text{moles} \]
- Using the stoichiometric ratio: \[ \text{O}_2 , \text{needed} = 17.19 , \text{moles of butane} \times 6.5 , \text{moles of O}_2/\text{mole of butane} \approx 111.73 , \text{moles of O}_2 \]
- Mass of O2 required: \[ \text{Mass of O}_2 = 111.73 , \text{moles} \times 32 , \text{g/mol} \approx 3575.36 , \text{g} = 3.576 , \text{kg} \]
Thus, approximately 3.576 kg of liquid oxygen is required for 1 kg of butane.
2. Moles of O2 Needed for 1 Mole of Alkanol
Assuming "alkanol" refers generically to alcohols like ethanol \((\text{C}_2\text{H}_5\text{OH})\) or others, the combustion can be described as: \[ \text{C}n\text{H}{2n+1}\text{OH} + O_2 \rightarrow CO_2 + H_2O \]
For complete combustion of ethanol (as an example): \[ \text{C}_2\text{H}_5\text{OH} + 3 O_2 \rightarrow 2 CO_2 + 3 H_2O \] Thus, you need 3 moles of \(O_2\) for each mole of ethanol.
3. Moles of H2O Formed for Each Mole of O2 Consumed
From the ethanol combustion: For every mole of \(O_2\) consumed, 1.0 mole of \(H_2O\) is gained (3 \(H_2O\)/3 \(O_2\)), hence there is a 1:1 ratio of moles of water to moles of oxygen consumed.
4. Grams of CO2 Produced for Each Mole of Alkanol Burned
The balanced reaction for ethanol combustion shows: \[ 2 \text{CO}_2\text{ formed} \text{ for every 1 mole of ethanol burned.} \]
Calculating the mass of CO2: \[ \text{Molar mass of } CO_2 = 44.01 , \text{g/mol} \] So for each mole of ethanol combusted, roughly 88.02 grams of \(CO_2\) is produced.
5. Grams of CO2 Produced for Each Gram of Alkanol Burned
The molar mass of ethanol (as an example): \[ \text{Molar mass of } \text{C}_2\text{H}_5\text{OH} = (2 \times 12.01) + (6 \times 1.008) + (16) = 46.08 , \text{g/mol} \]
Since 88.02 grams of CO2 is produced per 46.08 grams of ethanol, \[ \text{Grams of CO}_2 \text{ produced per gram of ethanol} = \frac{88.02 , \text{g}}{46.08 , \text{g}} \approx 1.91 , \text{g/g} \]
6. Mass of Copper Deposited from Zinc Reaction
From the equation \( \text{Zn} + \text{CuSO}_4 \rightarrow \text{ZnSO}_4 + \text{Cu} \):
1 mole of Zn produces 1 mole of Cu.
Calculate the moles of Zn: \[ \text{Moles of Zn} = \frac{6.5, \text{g}}{65.38 , \text{g/mol}} \approx 0.099 , \text{moles} \]
Thus, moles of Cu produced is also 0.099 moles: \[ \text{Molar mass of Cu} = 63.55 , \text{g/mol} \]
Mass of Cu produced: \[ \text{Mass of Cu} = 0.099 \times 63.55 \approx 6.30 , \text{g} \]
In conclusion:
- Mass of copper deposited = approximately 6.30 grams.
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