Can someone please help me?! Will award 10 points, thanks!?

1. 2Fe + 3H₂S0₄ → Fe₂(SO₄)₃ + 3H₂

2. 2C₂H₆ + 7O₂ → 6H₂O + 4CO₂

3. 3KOH + H₃PO₄ → K₃PO₄ + 3H₂O

4. SnO₂ + 2H₂ → Sn + 2H₂O

5. 4NH₃ + 5O₂ → 4NO + 6H₂O

6. 2KNO₃ + H₂CO₃ → K₂CO₃ + 2HNO₃

7. B₂Br₆ + 6HNO₃ → 2B(NO₃)₃ + 6HBr

8. 2BF₃ + 3Li₂SO₃ → B₂(SO₃)₃ + 6LiF

9. 4(NH₄)₃PO₄ + 3Pb(NO₃)₄ → Pb₃(PO₄)₄ + 12NH₄NO₃

10. SeCl₆ + O₂ → SeO₂ + 3Cl₂

Someone is pulling your leg, or they didn't bother to do their homework, or they just don't know any better. Several of the "reactions" you have been asked to balance simply do not occur or contain "compounds" that do not exist.

1. Fe + H2S04 > Fe2(SO4)3 + H2
The reaction of iron metal and dilute sulfuric acid produces iron(II) sulfate, not iron(III) sulfate. The correct equation is:
Fe(s) + H2SO4(aq) --> FeSO4(aq) + H2(g)

2. C2H6 + O2 > H2O + CO2
2C2H6(g) + 7O2(g) --> 6H2O + 4CO2

3. KOH + H3PO4 > K3PO4 + H2O
K3PO4 only occurs if there is a substantial excess of the base, otherwise KH2PO4 or K2HPO4 are the preferred products.
KOH(aq) + H3PO4(aq) --> KH2PO4(aq) + HOH(l)

4. SnO2 + H2 > Sn + H2O
SnO2(s) + 2H2(g) --> Sn(s) + 2H2O(g)

5. NH3 + O2 > NO + H2O
4NH3(g) + 3O2(g) --> 2N2(g) + 6H2O(g)

6. KNO3 + H2CO3 > K2CO3 + HNO3
There is no such molecule as H2CO3 in aqueous solution. It simply does not exist. What we call "carbonic acid" is actually a solution of carbon dioxide in equilibrium with minute amounts of H+ ion and HCO3- ions. Adding potassium nitrate to as solution of carbon dioxide will not produce a reaction.
KNO3(aq) + CO2(g) --> no reaction
Even what you have written does not indicate that a reaction has occurred. Both K2CO3 and HNO3 are soluble in water.
KNO3(aq) + "H2CO3" --> no reaction

7. B2Br6 + HNO3 > B(NO3)3 + HBr
There is no such compound as B(NO3)3. Boron does not behave as an alkali metal. While boron tribromide exists, there is no evidence of the dimer.

8. BF3 + Li2SO3 > B2(SO3)3 + LiF
As for the question above, there is no such compound as "B2(SO3)3". There is also little to suggest that "lithium sulfite" exists.

9. (NH4)3PO4 + Pb(NO3)4 > Pb3(PO4)4 + NH4NO3
Lead(IV) compounds are relatively rare and are usually quite simple. There is no lead(IV) nitrate, nor is there any lead(IV) phosphate. They simply do not exist.

10. SeCl6 + O2 > SeO2 + Cl2
There exists selenium tetrachloride and selenium hexafluoride, but there is no "selenium hexachloride."

Fe + H2S04 > Fe2(SO4)3 + H2
Balance the SO4 ions first

Fe + 3H2S04 > Fe2(SO4)3 + H2 now balance the H's
Fe + 3H2S04 > Fe2(SO4)3 +3 H2 now balance the Fe's

2Fe + 3H2S04 > Fe2(SO4)3 +3 H2

can one of the people who answered please help with one of my questions?

1. 2Fe + 3H2S04 > Fe2(SO4)3 + 3H2
2. 2C2H6 + 7O2 > 6H2O + 4CO2
3. 3KOH + H3PO4 > K3PO4 + 3H2O
4. SnO2 + 2H2 > Sn + 2H2O
5. 4NH3 + 5O2 > 4NO + 6H2O
6. 2KNO3 + H2CO3 > K2CO3 + 2HNO3
7. B2Br6 + 6HNO3 > 2B(NO3)3 + 6HBr
8. 2BF3 + 3Li2SO3 > B2(SO3)3 + 6LiF
9. 4(NH4)3PO4 + 3Pb(NO3)4 > Pb3(PO4)4 + 12NH4NO3
10. SeCl6 + O2 > SeO2 + 3Cl2

Done!!!!