439.3 g CO2
Explanation:
First find the # of moles of CO2 that results from the combustion of 3.327 mol C3H6:
3.227 mol C3H6 × (6 mol CO2/2 mol C3H6)
= 9.981 mol CO2
Use the molar mass of CO2 to determine the # of grams of CO2:
9.981 mol CO2 x (44.01 g CO2/1 mol CO2)
= 439.3 g CO2
Conc = moles / Vol you have C but need to find moles
moles KCl = mass / molar mass
25 g / 74.55 g,p;
0.335 moles
So rearrange the formula to find volume. Vol = moles / Conc
0.335 moles / 0.750 mol / L = 0.447 L or 447 mL
The answer in this question is 447 mL.
