Respuesta :
Answer:
No mass of butane could be left over since oxygen gas was in excess
Explanation:
Butane reacts with oxygen according to the equation;
2C₄H₁₀(g) + 13O₂(g) → 8CO₂(g) + 10H₂O(g)
This means every two moles of butane reacts with 13 moles of oxygen to form 8 moles of CO₂ and 10 moles of gaseous water.
We need to calculate mass of butane that remains;
Step 1: Moles of Butane and oxygen
Number of moles = Mass/ molar mass
Number of moles of Oxygen = 15.0 g/32 g/mol
= 0.46875 moles
Moles of butane in 2.3 g = 2.3 g / 58.12 g/mol
= 0.03957 moles
Step 2: Moles of butane required by 0.46875 moles of Oxygen
The mole ratio of butane to Oxygen is 2 : 13
Therefore; moles of butane will be;
= (0.46875/13)× 2
= 0.072115 moles
But the only amount of butane available is 0.03957 moles
This means the amount of oxygen available was in excess
Therefore; there would be no mass of butane over during the reaction since the reaction was in excess.
Answer:
The minimum mass that could be left over by the chemical reaction is 6.8 g.
Explanation:
The balanced chemical equation is
[tex]2 C_4 H_{10}+13 O_2>8 CO_2+10 H_2 O[/tex]
Since the quantity of two reactants are given using both we find the mass of the product.
The least mass is from the limiting reactant.
The more mass is obtained from the excess reactant.
Limiting reactant gets consumed first.
Excess reactant will be left over.
[tex]2.3 g C_{4} H_{10} \times \frac{1 m o l c_{4} H_{10}}{58 g C_{4} H_{10}} \times \frac{8 m o l C O_{2}}{2 m o l C_{4} H_{10}}=0.158 \mathrm{mol} \mathrm{CO}_{2}[/tex]
(LEAST)
[tex]15.0 \mathrm{g} \mathrm{O}_2 \times \frac{1 \mathrm{mol} \mathrm{O}_{2}}{32 \mathrm{gO}_{2}} \times \frac{8 \mathrm{mol} \mathrm{CO}_{2}}{13 \mathrm{mol} \mathrm{O}_{2}}=0.288 \mathrm{mol} \mathrm{CO}_{2}[/tex]
(EXCESS)
So, [tex]C_4 H_{10}[/tex] is the Limiting reactant and [tex]O_2[/tex] is the excess reactant
How much is excess? [tex]O_2[/tex] will be left over
[tex]2.3 g C_{4} H_{10} \times \frac{1 m o l C_{4} H_{10}}{58 g C_{4} H_{10}} \times \frac{13 \mathrm{mol} \mathrm{O}_{2}}{2 \mathrm{mol} C_{4} \mathrm{H}_{10}} \times \frac{32 \mathrm{gO}_{2}}{1 \mathrm{mol} \mathrm{O}_{2}}[/tex]
[tex]=8.23g O_2[/tex] is the actual need
So
[tex]15-8.23=6.8g O_2[/tex] is left unreacted (Answer)
