3C3H8 + 20H2CrO4-> 9CO2 + 10Cr2O3 + 32H2O

Question

Suppose you want to make 1.50 g of chromium (III) oxide, how many ml of 6.25M chromic acid will you need to use if the percent yield of the reaction is 68.6%?

Answer 1

You want 1.5 g Cr2O3 so the question you ask yourself is, "What number must I start with to get 1.5 g if the reaction is only 68.6% efficient?"
That's 0.686*X = 1.5 and X = 1.5/0.686 = about 2.3g or so but you should do this more accurately.
Then convert 2.3(actually the number you obtain) to mols.
mols Cr2O3 = 2.3g/molar mass Cr2O3.
Next, using the coefficients in the balanced equation, convert mols Cr2O3 to mols H2CrO4.
Finally, M H2CrO4 = mols H2CrO4/L H2CrO4. You have M and mols solve for L and convert to mL.

Answer 2

how do I convert moles of Cr2O3 to moles of H2CrO4? I don't understand what this means

Answer 3

Use the coefficients in the balanced equation.
?mols Cr2O3 x (20 mols H2CrO4/10 mols Cr2O3)