Answer:
The equation in option D is balanced
D) 8CO2 + 3S8 → 8CS2 + 8SO2
Explanation:
A) Pb(NO3)2 + FeCl3 → Fe(NO3)3 + PbCl2
⇒ This equation is <u>NOT balanced</u>
⇒ On the left side we have 3x Cl, while on the right side we have 2x Cl. To balance the amount of Cl on both sides we have to multiply FeCl3 (on the left side) by 2 and PbCl2 (on the right side) by 3
Pb(NO3)2 + 2FeCl3 → Fe(NO3)3 + 3PbCl2
⇒ now we have 2x Fe on the left side and 1x Fe on the right side. To balance the amount of Fe on both sides, we have to multiply Fe(NO3)3 (on the right side) by 2
Pb(NO3)2 + 2FeCl3 → 2Fe(NO3)3 + 3PbCl2
⇒ On the left we have 1x Pb , on the right side we have 3x Pb. To balance the amount of Pb on both sides we have to multiply Pb(NO3)2 (on the left side) by 3. Now the equation is balanced.
3Pb(NO3)2 + 2FeCl3 → 2Fe(NO3)3 + 3PbCl2
B) C3H5(NO3)3 → CO2 + H2O + N2 + O2
⇒ This equation is NOT balanced
⇒ On the left side we have 5x H, on the right side we have 2x H. To balance the amount of H on both sides we have to multiply C3H5(NO3)3 (on the left side) by 2 and H2O (on the right side) by 5
2C3H5(NO3)3 → CO2 + 5H2O + N2 + O2
⇒ On the left side we have 6x C, on the right side we have 1xC
To balance the amount of C on both sides we have to multiply CO2 (on the right side) by 6.
2C3H5(NO3)3 → 6CO2 + 5H2O + N2 + O2
⇒ On the left side we have 6x N, on the right side we have 2N
To balance the amount of N on both sides we have to multiply N2 (on the right side) by 3
.
2C3H5(NO3)3 → 6CO2 + 5H2O + 3N2 + O2
⇒ Now we have 18x O on the left side and 19x O on the right side
To balance the amount of O we should multiply 2C3H5(NO3)3 (on the left side) by 2 and multiply 6CO2 + 5H2O + 3N2 (on the right side) each by 2.
Now the equation is balanced.
4C3H5(NO3)3 → 12CO2 + 10H2O + 6N2 + O2
C) Fe2O3(s) + C(s) → Fe(S) + CO2(g)
⇒ This equation is NOT balanced
⇒ On the left side we have 2x Fe, on the right side we have 1x Fe. To balance the amount of Fe on both sides, we have to multiply Fe (on the right side) by 2
Fe2O3(s) + C(s) →2Fe(S) + CO2(g)
⇒ On the left side we have 3x O, on the right side we have 2x O
To balance the amount of O on both sides. we have to mulitply Fe2O3 (on the left side) by 2 and CO2 (on the right side) by 3. Then we see that we have 4x Fe on the left side, so we have to multiply 2FE (on the right side) by 2 as well.
2 Fe2O3(s) + C(s) → 4Fe(S) + 3CO2(g)
⇒ On the right side we 3x C, on the left side we have 1x C. To balance the amount of C on both sides, we have to multiply C (on the left side) by 3. Now the equation is balanced.
2 Fe2O3(s) + 3C(s) → 4Fe(S) + 3CO2(g)
D) 8CO2 + 3S8 → 8CS2 + 8SO2
⇒ This equation is balanced !
