Balance the following redox reaction if it occurs in H2SO4. What are the coefficients in front of C3H8O2 and H2SO4 in the balanced reaction? C3H8O2(aq) + K2Cr2O7(aq) → C3H4O4(aq) + Cr2(SO4)3(aq)

Balance elements other than O and H. Add H2O to balance oxygen, balance hydrogen by adding protons (H+) and Balance the charge of each equation with electrons

C3H8O2(aq) + 2H2O(l) → C3H4O4(aq) + 8H(+) + 8e(-)

Cr2O7(2-) + 14H(+) + 6e(-) → 2Cr(+3) + 7H2O (aq)

Scale the reactions so that the electrons are equal. If not look for a common multiple. In this case is 24, 8*3=6*4. So for each equation multiply for the correct multiple. Then

3C3H8O2(aq) + 6H2O(l) → 3C3H4O4(aq) + 24H(+) + 24e(-)

4Cr2O7(2-) + 56H(+) + 24e(-) → 8Cr(+3) + 28H2O (aq)

Add the reactions and cancel out common terms and group the ions in compounds if u can.

3C3H8O2(aq) + 32H(+) + 4Cr2O7(2-) → 3C3H4O4(aq) + 4Cr2(SO4)3(aq) + 22H2O (aq)

We are in acidic medium so the H(+) must be from H2SO4. So 32H(+) is 16H2SO4 in reactives. Add SO4(2-) in products and balance

16H2SO4 +3C3H8O2(aq) + 4Cr2O7(2-) → 3C3H4O4(aq) 4Cr2(SO4)3(aq) + 22H2O (aq) + 4SO4(2-)

Then

16H2SO4 +3C3H8O2(aq) + 4Cr2O7(2-) → 3C3H4O4(aq) 4Cr2(SO4)3(aq) + 22H2O (aq) + + 2K2SO4 (aq)