1 Expert Answer
2C4H10 + 13O2 ==> 8CO2 + 10H2O Balanced equation moles C4H10 present = 23.2 g x 1 mole/58.12 g = 0.3992 moles C4H10 At STP 1 mole of gas = 22.4 liters, so .. 0.3992 moles C4H10 x 22.4 L/mole = 8.94 liters of C4H10 From the balanced equation, mole ratio of CO2 : C4H10 is 8 : 2 so to find moles or liters of CO2 use this ratio. Liters CO2 = 8.94 liters C4H10 x 8 liters CO2/2 liters C4H10 = 35.8 liters CO2 produced Volume of O2 required = 8.94 L C4H10 x 13 L O2/2 L C4H10 = 58.1 liters of O2 required
1 Expert Answer
2C4H10 + 13O2 ==> 8CO2 + 10H2O Balanced equation
moles C4H10 present = 23.2 g x 1 mole/58.12 g = 0.3992 moles C4H10
At STP 1 mole of gas = 22.4 liters, so ..
0.3992 moles C4H10 x 22.4 L/mole = 8.94 liters of C4H10
From the balanced equation, mole ratio of CO2 : C4H10 is 8 : 2 so to find moles or liters of CO2 use this ratio.
Liters CO2 = 8.94 liters C4H10 x 8 liters CO2/2 liters C4H10 = 35.8 liters CO2 produced
Volume of O2 required = 8.94 L C4H10 x 13 L O2/2 L C4H10 = 58.1 liters of O2 required
Kolhan university
Aaron H.
Chemistry 101
6 months, 2 weeks ago
We don’t have your requested question, but here is a suggested video that might help.
What is the volume (in litres) of oxygen at STP required for complete combustion of $32 \mathrm{~g}$ of $\mathrm{CH}_{4}$ ? (mol. wt of $\mathrm{CH}_{4}=16$ ) (a) $89.6$ (b) $189.6$ (c) $98.4$ (d) $169.5$
1
Expert Answer2C4H10 + 13O2 ==> 8CO2 + 10H2O Balanced equation
moles C4H10 present = 23.2 g x 1 mole/58.12 g = 0.3992 moles C4H10
At STP 1 mole of gas = 22.4 liters, so ..
0.3992 moles C4H10 x 22.4 L/mole = 8.94 liters of C4H10
From the balanced equation, mole ratio of CO2 : C4H10 is 8 : 2 so to find moles or liters of CO2 use this ratio.
Liters CO2 = 8.94 liters C4H10 x 8 liters CO2/2 liters C4H10 = 35.8 liters CO2 produced
Volume of O2 required = 8.94 L C4H10 x 13 L O2/2 L C4H10 = 58.1 liters of O2 required
Aaron H.
Chemistry 101
5 months ago
We don’t have your requested question, but here is a suggested video that might help.
Methanol, $\mathrm{CH}_{4} \mathrm{OH}$, is a clean-burning, easily handled fuel. It can be made by the direct reaction of $\mathrm{CO}$ and $\mathrm{H}_{2}$. $$ \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(\ell) $$ (a) Starting with a mixture of $12.0 \mathrm{~g} \mathrm{H}_{2}$ and $74.5 \mathrm{~g} \mathrm{CO}$, which is the limiting reactant? (b) What mass of the excess reactant, in grams, is left after reaction is complete? (c) What mass of methanol can be obtained, in theory?