C3H8+3O2--->3CO2+8H
Therefore for every 1:3 there are 3 Carbon dioxides that form. That means find the limiting reactant from the two reactants.
5.5g(1mole C3H8/44.03g of C3H8)=0.1249 moled of C3H8 and if for every one C3H8 we can form three CO2. We can assume 0.3747 miles of CO2 will be produced.
15g of O2(1 mole O2/32g of O2)=0.4685moles O2 and if for every three O2 we can produce three CO2 we may assume a 1:1 ratio.
This means C3H8 will be your limiting reactant. Therefore 0.3747 moles of CO2 will be produced.
0.3747 moles of CO2(48.01 g of CO2/1 mole of CO2)= 17.99 grams of CO2
The volume of the dry gas at stp is calculated as follows
calculate the number on moles by use of PV =nRT where n is the number of moles
n is therefore = Pv/RT
P = 0.930 atm
R(gas contant= 0.0821 L.atm/k.mol
V= 93ml to liters = 93/1000= 0.093L
T= 10 + 273.15 = 283.15k
n= (0.930 x0.093) /(0.0821 x283.15) = 3. 72 x10^-3 moles
At STp 1 mole = 22.4L
what about 3.72 x10^-3 moles
by cross multiplication
volume = (3.72 x10^-3)mole x 22.4L/ 1 moles = 0.083 L or 83.3 Ml
1. B
2. C
3. E
4. D
5. A
6. B
7. A
8. E
9. A
Hope this helps!
