To determine the stoichiometric coefficients for the given chemical reaction, you need to balance the equation by ensuring that the number of atoms on both sides of the reaction is the same. Here's a step-by-step guide:
1. Start by balancing the elements that appear in only one compound on each side of the equation. In this case, we have Cr, C, H, O, K, and S.
2. Balance the Cr atoms: On the left side, we have 2 Cr atoms, and on the right side, we have Cr2. To balance Cr, put a coefficient of 2 in front of Cr2(SO4)3 on the right side. This gives us:
K2Cr2O7 + H2SO4 + C2H5OH ---> 2Cr2(SO4)3 + K2SO4 + CH3COOH + H2O
3. Balance the C atoms: On the left side, we have 2 C atoms in C2H5OH, and on the right side, we have 1 C atom in CH3COOH. To balance C, put a coefficient of 2 in front of CH3COOH on the right side. This gives us:
K2Cr2O7 + H2SO4 + C2H5OH ---> 2Cr2(SO4)3 + K2SO4 + 2CH3COOH + H2O
4. Balance the H atoms: On the left side, we have 6 H atoms in C2H5OH and 2 H atoms in H2O, adding up to a total of 8 H atoms. On the right side, we have 4 H atoms in 2CH3COOH and 2 H atoms in H2O, adding up to a total of 6 H atoms. To balance H, put a coefficient of 4 in front of H2O on the left side. This gives us:
K2Cr2O7 + H2SO4 + C2H5OH ---> 2Cr2(SO4)3 + K2SO4 + 2CH3COOH + 4H2O
5. Balance the O atoms: On the left side, we have 7 O atoms in K2Cr2O7 and 4 O atoms in H2SO4, adding up to a total of 11 O atoms. On the right side, we have 24 O atoms in 2Cr2(SO4)3, 4 O atoms in K2SO4, 4 O atoms in 2CH3COOH, and 8 O atoms in 4H2O, adding up to a total of 40 O atoms. To balance O, put a coefficient of 12 in front of K2Cr2O7 on the left side. This gives us the final balanced equation:
12K2Cr2O7 + 3H2SO4 + 8C2H5OH ---> 24Cr2(SO4)3 + 4K2SO4 + 4CH3COOH + 8H2O
Therefore, the stoichiometric coefficients for the reaction are:
K2Cr2O7: 12
H2SO4: 3
C2H5OH: 8
Cr2(SO4)3: 24
K2SO4: 4
CH3COOH: 4
H2O: 8