Okay, in this prolem i need to balence this equation. Even our teacher had problems with it, but he refuses to help anyone.
PbCrO4 + HCL + FeSO4 ----> PbCl2 + Cr2 (SO4)3 + H2O + Fe2(SO4)3
please help me. this is the last problem on my homework.
then im off to start math :)
Thank you!
2PbCrO4 + HCl + 2FeSO4 ----> PbCl2 + Cr2 (SO4)3 + H2O + Fe2(SO4)3
In the above equation, I have adjusted two of the coefficients. Now, just go through and balanced the oxidation/reduction part. Keep the number of electrons equal.
Cr changes from +6 for each atom on the left to +3 for each atom on the right. Fe changes from +2 for each atom on the left to +3 for each atom on the right.
To balance the given chemical equation PbCrO4 + HCl + FeSO4 → PbCl2 + Cr2(SO4)3 + H2O + Fe2(SO4)3, we need to ensure that the number of atoms of each element is equal on both sides of the equation. Here's a step-by-step breakdown of how to balance the equation:
1. Start by balancing the atoms other than oxygen and hydrogen. In this case, we have lead (Pb), chromium (Cr), chlorine (Cl), sulfur (S), and iron (Fe).
PbCrO4 + HCl + FeSO4 → PbCl2 + Cr2(SO4)3 + H2O + Fe2(SO4)3
2. Balance the elements that appear in only one compound on each side of the equation first. In this case, we have lead (Pb) and chromium (Cr).
PbCrO4 + HCl + FeSO4 → PbCl2 + Cr2(SO4)3 + H2O + Fe2(SO4)3
3. To balance the number of Pb atoms, place a coefficient of 2 in front of PbCl2.
PbCrO4 + HCl + FeSO4 → 2PbCl2 + Cr2(SO4)3 + H2O + Fe2(SO4)3
4. Next, balance the number of Cr atoms. Place a coefficient of 3 in front of Cr2(SO4)3.
PbCrO4 + HCl + FeSO4 → 2PbCl2 + 3Cr2(SO4)3 + H2O + Fe2(SO4)3
5. Now, balance the number of Cl atoms. Place a coefficient of 6 in front of HCl.
PbCrO4 + 6HCl + FeSO4 → 2PbCl2 + 3Cr2(SO4)3 + H2O + Fe2(SO4)3
6. Next, balance the number of S atoms. Place a coefficient of 5 in front of FeSO4.
PbCrO4 + 6HCl + 5FeSO4 → 2PbCl2 + 3Cr2(SO4)3 + H2O + Fe2(SO4)3
7. Finally, balance the number of H and O atoms. At this point, all the elements should be balanced except for hydrogen and oxygen. To balance them, place a coefficient of 22 in front of H2O.
PbCrO4 + 6HCl + 5FeSO4 → 2PbCl2 + 3Cr2(SO4)3 + 22H2O + Fe2(SO4)3
Now, the equation is balanced with the same number of atoms on both sides. Double-check your balanced equation to make sure all elements are balanced correctly.