Need help in Chemistry homework- Balancing Chemical Equations?

1) Na+ H2O ---->NaOH +1/2 H2

2) HNO3+NaHCO3 ----------> NaNO3+ H2O + CO2 it is already balanced

3) H2O+ 1/2O2 ------------> H2O2

1) 2 Na + 2 H2O -> 2 NaOH + H2

2) seems like it's already balanced :)

3) 2 H2O + O2 -> 2 H2O2

hope this helps.. not entirely sure if i'm right though :P

because a number of those reactions are redox, the balancing can get problematical. a million. it might help to bear in mind that SO4= is an ionic crew. So, each and every of the Pb for PbSO4 comes from both lead entities at the same time as the SO4= comes from the sulfuric acid. obviously, there'll be more suitable than a million unit of H2SO4 in touch. If we double the H2SO4 and the PbSO4, the Pb and SO4 are in stability. yet we've 4H left from the H2SO4 and 2O from the PbO2. That makes 2H2O, so as that is cool. Then..... 2H2SO4 + Pb + PbO2 -> 2PbSO4 + 2H2O 2. each and every of the carbon comes from the sugar to boot as each and every of the hydrogen for water. the different oxygen source balances out the shortfall in starting to be oxygen for CO2 and water. So we would have 6CO2 and 6H2O. we favor 18O for those compounds, in spite of the undeniable fact that the sugar in user-friendly words provides you 6, so the different 12O are offered interior the O2. Then....... C6H12O6 + 6O2 -> 6CO2 + 6 H2O 3. This sounds like an "each and every guy for himself" stability challenge. So shall we from from left to accurate. a million unit of reagent calcium phosphate will provide sufficient Ca for 3 instruments of SiO3. If we make that adjustment, there are 2P on the left area and four on the right. so that you would possibly want to provide sufficient P for P4, we favor to double the calcium phosphate, which doubles the calcium silicate from the previous argument. At this element, we've (no longer completely performed)....... 2 Ca3(PO4)2 + C + SiO2 -> 6CaSiO3 + CO +P4 The Ca and P are chuffed, yet we favor to multiply the SiO2 on the left by way of 6 to provide Si for the calcium silicate. Then...... 2 Ca3(PO4)2 + C + 6 SiO2 -> 6 CaSiO3 + CO + P4 This brings us to the oxygen. There are 28 instruments on the left and 19 on the right. yet we may be able to toss in sufficient C to create more suitable CO and make up the oxygen deficit on the right. Then we've........ 2 Ca3(PO4)2 + 10C + 6SiO2 -> 6 CaSiO3 + 10CO + P4. 4. because acetate ion isn't altered, this seems a lot less daunting in case you enable that ion be represented by way of a few thing like "X". Then.... HX + NaHCO3 -> CO2 + NaX + H2O. properly,............

2Na + 2H2O-->2NaOH + H2

HNO3 + NaHCO3---> NaNO3 + H2O + CO2

2H2O + O2---> 2H2O2