For the reaction of combustion of 74.3 g of butane with 132 kg of oxygen, we have:
a. The limiting reactant is butane
b. The mass of carbon dioxide obtained by heating butane is 225.33 grams.
c. 131734.8 grams or 131.73 kilograms of oxygen remains unreacted.
a. The balanced reaction is the following:
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O (1)
To find the limiting reactant, we need to calculate the initial number of moles of butane and oxygen
[tex] n_{C_{4}H_{10}}_{i} = \frac{m_{C_{4}H_{10}}}{M_{C_{4}H_{10}}} = \frac{74.3 g}{58.12 g/mol} = 1.28 \:moles [/tex]
[tex] n_{O_{2}}_{i} = \frac{m_{O_{2}}}{M_{O_{2}}} = \frac{132000 g}{31.99 g/mol} = 4126.3 \:moles [/tex]
From equation (1) we have that 2 moles of C₄H₁₀ react with 13 moles of O₂, so the number of oxygen moles needed to react with 1.28 moles of C₄H₁₀ is:
[tex] n_{O_{2}}_{r} = \frac{13\:moles\:O_{2}}{2\:moles\:C_{4}H_{10}}*1.28\:moles\:C_{4}H_{10} = 8.32\:moles [/tex]
We need 8.32 moles of O₂ to react with butane and initially we have 4126.3 moles, so the limiting reactant is butane.
b. The mass of CO₂ can be found with the limiting reactant
From equation (1), 2 moles of C₄H₁₀ react with oxygen to produce 8 moles of CO₂, so the number of moles of CO₂ produced is:
[tex] n_{CO_{2}} = \frac{8\:moles\:CO_{2}}{2\:moles\:C_{4}H_{10}}*1.28\:moles\:C_{4}H_{10} = 5.12\:moles [/tex]
Then, the mass of CO₂ is:
[tex] m_{CO_{2}} = n_{CO_{2}}*M_{CO_{2}} = 5.12 \:moles*44.01 g/mol = 225.33 g [/tex]
Hence, the mass of CO₂ obtained is 225.33 grams.
c. The number of moles of the excess reagent can be calculated as follows
[tex] n_{O_{2}} = n_{O_{2}}_{i} - n_{O_{2}}_{r} = (4126.3 - 8.32) \:moles = 4118 \:moles [/tex]
Now, the mass of oxygen that remains after the reaction is:
[tex] m_{O_{2}} = 4118\:moles*31.99g/mol = 131734.8 g = 131.73 kg [/tex]
Therefore, 131734.8 g or 131.73 kg of oxygen remains unreacted.
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