How do I balance this chemical equation CH3COOH+NaHCO3=CO2+H2O+NaCO3?
551370
Well, the equation is..\nCH3COOH + NaHCO3 ----> CH3COONa (Sodium Acetate) + H2CO3 (Carbonic Acid) \n(or)\nCH3COOH + NaHCO3 ----> CH3COONa (Sodium Acetate) + H2O (Water) + CO2 (Carbon-dioxide)\n\nSo you might have noticed the equation is already balanced. It is a double-displacement reaction. So the sides are already balanced.\nHope this helps.\n\nOne small note...I don't think there is a compound such as NaCO3. The valency of CO3(^-2) ion is 2, so it has to combine with two Sodiums (two positive charges to neutralise its two negative charges) unless the compound one of the above users mentioned is NaCO3-.\n\nrule wid style n luv@vinny
802803
The answer is:\n2Al(HCO3)3  -------> Al2(CO3)3  +  3CO2  +  3H20\n\nThe equation is now fully balanced as you have\n2 moles of Al on each side\n6 moles of H on each side\n6 moles of C on each side\n18 moles of O on each side\n\nHope that helps!
758773
6O2 + C6H12O6 ---> 6CO2 + 6H2O
577084
Acid + base = salt + water.\n\nUnbalanced eqn is: Al(OH)3 + H2CO3 --> Al2(CO3)3 + H2O
842372
Write the balanced chemical equation:\nNaOH + HNO3 -----> H2O + NaNO3\n\nthis means that 1 mole of NaOH is required to use up 1 mole of HNO3\nthis means that all of your HNO3 will be used up so to determine the excess NaOH all you have to do is subtract the # of moles of HNO3 from the number of moles of NaOH giving 0.00193 moles NaOH left.
745644
H3PO4 + 3NaOH ----> 3H2O + Na3PO4
1070616
mix concetrated sulphuric acid to the sodium acetate, the reaction will produce acetic acid and sodium sulphate.\n\nif you take all the water out of the sodium acetate, you will end up with 100% pure acetic acid.\n\nH2SO4 + CH2COONa => CH3COOH + Na2SO4\n(unbalanced equation)
1097630
Na2CO3 + 2HCl-> NaCl + H2O + CO2
425080
NaCO3(s) -->  2Na+(aq)  + CO3 2-(aq)
904363
(NH4)2CO3+MgSO4->(NH4)2SO4+MgCO3
50281
This is an acid base reaction, but the other answer is incorrect about the product. There is no Cl present. Here’s the balanced equation:\n\n2NaOH (s) + H2CO3 (aq) ➔ Na2CO3 + 2H2O\n\nNaOH is the base. H2CO3 is the acid.
907209
Na2CO3 + NaHCO3  \n\nmoles Na2CO3 = 0.070L * 5.5 mol/L = 0.385 mol\nmoles of Na+ from Na2CO3 = 0.77 mol\nmoles NaHCO3 = 0.030L * 1.0 mol/L = 0.030 mol\nmoles of Na+ from NaHCO3 = 0.03 mol\ntotal moles of Na+ in the solution = 0.80 mol\nsodium ion (Na+) concentration = 0.80 mol/0.1L = 8.0M
122744
Mg + 2 CH3COOH = Mg(CH3COO)2 + H2
930806
When you want to balance ANY chemical equation, always follow these steps:\n\n1. Balance metals\n2. Balance non-metals\n3. Balance hydrogen\n4. Balance oxygen\n\nLet's do it\n\nFirst let's balance K (metal):\n\n"2" KHCO3 + __H3PO4 yields __K2HPO4 + __H2O + __CO2\n\ndone. now let's balance P (non metal): if you check, it is balanced already.\n\nnow C (also non metal):\n\n2 KHCO3 + __H3PO4 yields __K2HPO4 + __H2O + "2" CO2\n\nnow hydrogen:\n\n2 KHCO3 + __H3PO4 yields __K2HPO4 + "2" H2O + 2 CO2\n\nfinally, oxygen (its already balanced!!)\n\nSo you end with\n\n2 KHCO3 + H3PO4 yields K2HPO4 + 2 H2O + 2 CO2
729249
first balance equation\nCH4 + 2O2 --> CO2 + 2H2O\n\nthen convert (look at balanced equation and use coefficients)\n3.28 moles O2 x 1 Mole of CO2 / 2 Moles O2 = 1.64 moles of CO2
462663
C4H9OH + 6O2 --> 4CO2 + 5H2O
700313
NaOH + K+ = Na+ + KOH