Need help with balancing these equations....? plzz?

Question

1.) Na + H2O ----> NaOH + H2
2.) Na2B4O7 + HCl + H2O ----> H3BO3 + NaCl
3.) Fe2(SO4)3 + Ba(OH)2 ----> Fe(OH)3 + BaSO4
4.) C12H22O11 + H2O ----> C2H5OH + CO2


i just cant balance them. all the numbers are subscripts

Answer 1

its very simple
1. 2Na + 2H2O = 2NaOH + H2
2. Na2B4O7 + 2HCl + 5H2O = 4H3BO3 + 2NaCl
3. Fe2(SO4)3 + 3Ba(OH)2 = 2Fe(OH)3 + 3BaSO4
4. C12H22O11 + H2O = 4C2H5OH + 4CO2

Answer 2

1) 2Na + 2H2O = 2NaOH + H2
2) Na2B4O7 + 2HCl + 5H20 = 4H3BO3 + 2NaCl
3) Fe2(SO4)3 + 3Ba(OH)2 = 2Fe(OH)3 + 3BaSO4
4) C12H22O11 + H2O = 4C2H5OH + 4CO2

That's what I got. Take it from Axl Rose; " Don't give up Sweet Child O mine. "
What you need to do is to notice how many of each of the elements there are on both sides of the equation. in number 2. there are 2 Sodium atoms on the left. So,there must also be 2 on the right. etc.
Most importantly; practice makes perfect. Or at least acceptable. No body's perfect. We all make mistakes. But, Never give up.

Answer 3

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

Na2B4O7 + 2HCl + 5H2O ----> 4H3BO3 + 2NaCl
(Balance from the number of borons)

Fe2(SO4)3 + 3Ba(OH)2 ----> 2Fe(OH)3 + 3BaSO4
(Balance from the iron (III) sulfate)

C12H22O11 + H2O ----> 4C2H5OH + 4CO2
(Fermentation equation each alcohol molecule forms with a carbon dioxide molecule also) ( I divided the 3 carbons on the right into the number of carbons (12) on the left to start with four (4) of each product and voila it was balanced.

Check my work.......:>)

Answer 4

1. 2Na + 2H2O ~~~> 2NaOH + H2
2. Na2B4O7 + 2HCl + 5H2O ~~~> 4H3BO3 + 2NaCl
3. Fe2(SO4)3 + 3Ba(OH)2 ~~~> 2Fe(OH)3 + 3BaSO4
4. C12H22O11 + H2O ~~~> 4C2H5OH + 4CO2

Answer 5

1.) 2Na + 2H2O ----> 2NaOH + H2
2.) Na2B4O7 + 2HCl + 5H2O ----> 4H3BO3 + 2NaCl
3.) Fe2(SO4)3 + 3Ba(OH)2 ----> 2Fe(OH)3 + 3BaSO4
4.) C12H22O11 + H2O ----> 4C2H5OH + 4CO2

Answer 6

1)2Na + 2H2O --> 2NaOH + H2
2)Na2B4O7 + 2HCl + 5H2O --> 4H3BO3 + 2NaCl
3)Fe2(SO4)3 + 3Ba(OH)2 --> 2Fe(OH)3 + 3BaSO4
4)C12H22O11 + 2H2O --> 4C2H5OH + 4CO2

Answer 7

2Na + 2H2O ===> 2NaOH + H2

Na2B4O7 + 2HCl + 5H2O ===> 4H3BO3 + 2NaCl

Fe2(SO4)3 + 3Ba(OH)2 ===> 2Fe(OH)3 + 3BaSO4

C12H22O11 + H2O ===> 4C2H5OH + 4CO2

Answer 8

1) C2H2O4 is ethandioic (oxalic) acid, also written as (COOH)2 or H2C2O4 and dissociating into 2H^+ and C2O4^2- (ethandioate/oxalate) ions which are oxidised to CO2 a) C2O4^2- ----->2CO2 b) C(lll) -------> C(lV) + e Work out oxidation states of C before and after and number of e's to alter it c) C2O4^2- -----> 2CO2 + 2e Insert the whole species involved and correct number of e's d) C2O4^2- -----> 2CO2 + 2e Check balancing; 4O, 2C AND 2- CHARGE on each side Repeat steps a) to c) then the further steps a) MnO4^- --------> Mn^2+ b) Mn(VII) + 5e ------> Mn(II) c) MnO4^- + 5e --------> Mn^2+ d) MnO4^- + 5e --------> Mn^2+ + 4H2O Balance O by including appropriate number of H2O moles e) 8H^+ + MnO4^- + 5e --------> Mn^2+ + 4H2O Balance H by including appropriate number of H^+ Check balancing; 8H, Mn, 4O AND OVERALL 2+ on each side (8+ +1- + 5- = 2+ on left) Combine two half equations by 5xd) + 2xe) which eliminates e's (10 on each side) 16H^+ + 2MnO4^- + 5C2O4^2- -----> 10CO2 + 2Mn^2+ + 8H2O Not essential but 10 of the H^+ can be joined to the oxalate ions to give the acids formula but notice the reaction still needs 6H^+ ie. acidic conditions 6H^+ + 2MnO4^- + 5H2C2O4 -----> 10CO2 + 2Mn^2+ + 8H2O Final balance check; 16H, 2Mn, 28O 10C AND 4+ CHARGE on each side 2) a) (you couldn't have a n ion-electron half equation much simpler than this!!!) a) NO3^- -------> NH4^+ This one will not be so simple! Repeat the a), b), etc steps as far as needed b) N(V) ie 5+ + 8e -------> N(lll-) ie 3- c) NO3^- + 8e -------> NH4^+ d) NO3^- + 8e -------> NH4^+ 3H2O e) 10H^+ + NO3^- + 8e -------> NH4^+ + 3H2O Balancing check; 10H, N, 3O and one + overall on each side To combine the two ion-electron half equation and eliminate e's 4xfirst + second 10H^+ + NO3^- + 4Sn -------> 4Sn^2+ + NH4^+ + 3H2O Final balance check; 10H, N, 3O, 4Sn and the all important 9+ charge, on each side I have explained quite a lot but must assume you have reached a level where you are competent at working out oxidation states in complex ions eg S(Vl) or S(6+) in SO4^2-. Also that you can accurately count up atoms on each side of an equation AND OVERALL CHARGES.

Answer 9

You're thinking whole numbers only (thats why you CAN'T balance them).

look at the first one as :
Na + H20 ----> NaOH + 1/2 H2 then:
2Na + 2 H20 -----> 2 NaOH + H2

Answer 10

This is just a comment to the guy about whole numbers--you can't have half an atom. That's why they have to be in whole numbers.