To solve this we use the equation,
M1V1 = M2V2
where M1 is the concentration of the stock
solution, V1 is the volume of the stock solution, M2 is the concentration of
the new solution and V2 is its volume.
65 x V1 = 2 x 200 L
V1 = 6.15 L
Is this the whole answer?
The first step is to balance the equation:
<span>C3H8 + 5O2 ---> 3CO2 + 4H2O
Check the balance
element left side right side
C 3 3
H 8 4*2 = 8
O 5*2=10 3*2 + 4 = 10
Then you have the molar ratios:
3 mol C3H8 : 5 mol O2 : 3 mol CO2 : 4 mol H2O
Now you have 40 moles of O2 so you make the proportion:
40.0 mol O2 * [3 mol CO2 / 5 mol O2] = 24.0 mol CO2.
Answer: option D. 24.0 mol CO2
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