10)
a.
-calculate the number of moles of butane [n=m/MM, n=65/(1.008*10) +(12.01*4) ---- n=65/58.12 = 1.118...]- you are dividing the mass of butane by its molar mass 58.12.
-then calculate the number of moles of oxygen- as it is at SLC, then as 1 mol=24.8L, that means that 200L=8.06...mol (200/24.8 = 8.06...)
Then, use stoichiometric ratios:
I'll use the number of moles of butane as a reference. The molar ratio of oxygen to butane is 13/2, so 1.118...*7.5 =7.2694....
So, butane is the limiting reagent, because according to the calculations oxygen is in excess.
b.
as the number of moles of the limiting reagent is 1.118..., use that value and the stoichiometric ratios to calculate the number of moles of CO2. So, 1.118... * (8/2)= 4.4735...moles. Convert it to litres: 4.4735...moles * 24.8 L = 111 L (3 sig figs).
