can you give me some equestions for balance equation for 10th std
Answers
Answer:
you do relaxing. no more thoughts about it. just do work for your study.and developed excellence
answer:-yes ...
here r some equations for u..
FeCl3 + MgO ---> Fe2O3 + MgCl2
Solution:
1) Balance the Cl (note that 2 x 3 = 3 x 2):
2FeCl3 + MgO ---> Fe2O3 + 3MgCl2
The Fe also gets balanced in this step.
2) Pick either the O or the Mg to balance next:
2FeCl3 + 3MgO ---> Fe2O3 + 3MgCl2
The other element (Mg or O, depending on which one you picked) also gets balanced in this step.
Problem #2: Li + H3PO4 ---> H2 + Li3PO4
Solution:
1) Balance the Li:
3Li + H3PO4 ---> H2 + Li3PO4
2) Now, look at the hydrogens. See how the H comes only in groups of 3 on the left and only in groups of 2 on the right? Do this:
3Li + 2H3PO4 ---> 3H2 + Li3PO4
Remember 2 x 3 = 6 and 3 x 2 = 6. It shows up a lot in balancing problems (if you haven't already figured that out!).
3) Balance the phosphate as a group:
3Li + 2H3PO4 ---> 3H2 + 2Li3PO4
4) Oops, that messed up the lithium, so we fix it:
6Li + 2H3PO4 ---> 3H2 + 2Li3PO4
Problem #3: ZnS + O2 ---> ZnO + SO2
1) Balance the oxygen with a fractional coefficient (Zn and S are already balanced):
ZnS + (3/2)O2 ---> ZnO + SO2
2) Multiply through to clear the fraction:
2ZnS + 3O2 ---> 2ZnO + 2SO2
Problem #4: FeS2 + Cl2 ---> FeCl3 + S2Cl2
Solution:
See how the Fe and the S are already balanced? So, look just at the Cl. There are a total of 5 on the right-hand side, so we put 5 on the left:
FeS2 + (5/2)Cl2 ---> FeCl3 + S2Cl2
Clear the fraction by multiplying through by 2:
2FeS2 + 5Cl2 ---> 2FeCl3 + 2S2Cl2
Problem #5: Fe + HC2H3O2 ---> Fe(C2H3O2)3 + H2
Solution:
1) Balance the acetate:
Fe + 3HC2H3O2 ---> Fe(C2H3O2)3 + H2
2) Balance the hydrogen:
Fe + 3HC2H3O2 ---> Fe(C2H3O2)3 + (3/2)H2
3) Clear the fraction:
2Fe + 6HC2H3O2 ---> 2Fe(C2H3O2)3 + 3H2
Problem #6: H2(g) + V2O5(s) ---> V2O3(s) + H2O(ℓ)
Solution:
1) Balance the oxygen:
H2(g) + V2O5(s) ---> V2O3(s) + 2H2O(ℓ)
2) Balance the hydrogen:
2H2(g) + V2O5(s) ---> V2O3(s) + 2H2O(ℓ)
Note that the vanadium was not addressed because it stayed in balance the entire time. Note how the hydrogen started out balanced, but the balancing of oxygen affected the hydrogen, which we addressed in the second step.
Problem #7: HCl(aq) + MnO2(s) ---> MnCl2(aq) + Cl2(g) + H2O(ℓ)
Solution:
1) Balance the chlorine:
4HCl(aq) + MnO2(s) ---> MnCl2(aq) + Cl2(g) + H2O(ℓ)
2) Balance the hydrogen:
4HCl(aq) + MnO2(s) ---> MnCl2(aq) + Cl2(g) + 2H2O(ℓ)
With this last step, the oxygen is also balanced and the Mn was never mentioned because it started out balanced and stayed that way.
Problem #8: Fe2O3(s) + C(s) ---> Fe(s) + CO2(g)
Solution:
1) Balance the iron:
Fe2O3(s) + C(s) ---> 2Fe(s) + CO2(g)
2) Balance the oxygen:
Fe2O3(s) + C(s) ---> 2Fe(s) + 3⁄2CO2(g)
3) Balance the carbon:
Fe2O3(s) + 3⁄2C(s) ---> 2Fe(s) + 3⁄2CO2(g)
Note the 3⁄2 in front of the C and the CO2. What's that you say? You can't have 3⁄2 of an atom? Ah, just you wait.
4) Multiply through by two for the final answer:
2Fe2O3(s) + 3C(s) ---> 4Fe(s) + 3CO2(g)
Comment: one way to look at this is that using the 3⁄2 was just a mathematical artifice to balance the equation. The chemical reality of atoms reacting in ratios of small whole numbers is reflected in the final answer.
Another way to look at the coefficients is in terms of moles. We can certainly have 3⁄2 of a mole of carbon atoms or 3⁄2 of a mole of carbon dioxide molecules. The final step towards whole number coefficients is just a convention. The chemical equation is balanced in a chemically-correct sense with the fractional coefficients.
Problem #9: C5H11NH2 + O2 ---> CO2 + H2O + NO2
Solution:
1) Balance the hydrogens first:
2C5H11NH2 + O2 ---> CO2 + 13H2O + NO2
Notice that I used a 2 in front of C5H11NH2. That's because I knew that there are 13 hydrogens in the C5H11NH2 and that meant a 13⁄2 in front of the water. I knew I'd have to eventually clear the 13⁄2, so I decided to do so right at the start.
2) Balance the nitrogen and the carbon:
2C5H11NH2 + O2 ---> 10CO2 + 13H2O + 2NO2
3) Oxygen:
2C5H11NH2 + 37⁄2O2 ---> 10CO2 + 13H2O + 2NO2
4) Multiply through by 2 for:
4, 37 ---> 20, 26, 4
Problem #10: CO2 + S8 ---> CS2 + SO2
Solution:
1) The only thing not balanced already is the S:
CO2 + 3⁄8S8 ---> CS2 + SO2
Most of the time the fraction used to balance is something with a 2 in the denominator: 1⁄2 or 5⁄2 or 13⁄2, for example. Not too often does one see 3⁄8. Pretty tricky!
2) Multiply through by 8:
hope it helps u
8CO2 + 3S8 ---> 8CS2 + 8SO2