Balanced Equations?

Question

Please tell me what these turn into
Double replacement Reactions:
NaCl + AgNO3 turned into a solid
Na2CO3 + HCl fizzed
NH4Cl + NaOH ammonia present
BaCl2 + Na2SO4 milky appearance (turned into a solid)
Na2CO3 + H2SO4 precipitated
NaOH + FeSO4 Immediately turned brown

Single Replacement Reactions:
H2O + Ca Produced gas
HCl + Al completely disolved in 5 minutes
HCl + Zn produced gas
HCl + Cu solution turned very light yellow
CuCl2 + Zn turned black on edges and precipitated
AgNO3 + Cu flaking apart, oxidizing, corroding

Can you please balance these and let me know how YOU know what they turned into? Im lost with these equations. Thanks so much. Angel

Answer 1

-NaCl + AgNO3 -> NaNO3 + AgCl (white solid. insoluble)
-Na2CO3 + 2HCl ->2NaCl + H2O +CO2(fizzle gas from acid/base rxn)
-NH4Cl + NaOH -> NaCl + H2O +NH3 (NH4+ is conjugate acid of NH3, so it reacts with NaOH in acid/base rxn)
-BaCl2 + Na2SO4 ->2NaCl + BaSO4(white solid. insoluble)
-Na2CO3 + H2SO4->Na2SO4+H2O+CO2(i cant explain precipitation. sodium sulphate is pretty soluble, as are all sodium compounds.)
-2NaOH + FeSO4->Na2SO4 + Fe(OH)2 (yellow precipitate. oxidises in air quickly to form iron(iii)hydroxide which is brown)

-2H2O + Ca -> Ca(OH)2 + H2 (hydrogen gas)
-6HCl + 2Al-->2AlCl3 + 3H2 (same)
-2HCl + Zn --->ZnCl2 +H2(same)
-4HCl + Cu --->[CuCl4]2- + 4H+ (copper complex ion which is yellow in color. pure copper will never react with HCl. requires pretty high conc of HCl for the complex ion to arise.)
-CuCl2 + Zn -->ZnCl2 + Cu (reddish copper metal precipitates from solution onto the zinc metal reactant. i have no explanation on how the black arises)
-2AgNO3 + Cu->Cu(NO3)2 + 2Ag(silver precipitates onto copper metal reactant, which might flake off when copper is used up. aqueous silver nitrate is an oxidising agent and is quite corrosive).

sorry i couldnt give an answer to all the questions.

Answer 2

Double replacement reactions:

NaCl + AgNO3 = NaNO3 + AgCl (solid)

Na2CO3 + 2HCl = 2NaCl + H20 +CO2 (gas)

NH4Cl + NaOH = NaCl + H2O + NH3 (ammonia)

BaCl2 + Na2SO4 = 2NaCl + BaSO4 (solid)

Na2CO3 + H2SO4 = Na2SO4 + H2O + CO2 (i'm not sure where you get the precipitate from. i don't think there shouldn't be any.)

if you last equation is correct, it's:
2NaOH + FeSO4 = Na2SO4 + Fe(OH)2 (the Fe2+ ion should cause it to turn dirty green 1st then turn brown because the Fe2+ ion gets oxidized to Fe3+ ion)

if it turns brown immediately, then the equation should be:
6NaOH + Fe2(SO4)3 = 3Na2SO4 + 2Fe(OH)3 (with the Fe3+ ion causing the brown precipitate)

you balance the equations by making sure that there are same number of each elements on both sides.
there is something like, an exchange of ions between both compounds. unfortunately, there's no easy way (as far as i know) to know which compounds exists as solids or not. the only method i know, is to memorise the information.

try the single replacement reactions yourself. don't expect others to do all your work for you. if you run into problems, refer to your textbooks or just ask your teachers for help. i'm sure your teachers won't be giving you assignments based on things that are not taught to you.

Answer 3

Dear Angel,

Here are my suggestions.

Sodium chloride plus silver nitrate will give a white precipitate of silver chloride:
Ag+(aq) + Cl-(aq) --> AgCl(s)

The fizzing is carbon dioxide:
CO3--(aq) + 2H+ --> CO2(g) + H2O(l)

This is simply NH4+(s) + OH-(s) --> NH3(g) + H2O

The white solid is barium sulphate:
Ba++(aq) + SO4--(aq) --> BaSO4(s)

I'm not too sure about the next one. Unless, that is , you started with a very concentrated solution of sodium carbonate. Then it is possible that sodium bicarbonate (sodiumn hydrogen carbonate), which is a lot less soluble than the carbonate itself, could precipitate: CO3--(aq) + H+(aq) --> HCO3-(s)

If you had started off with pure FeSO4, the precipitate should have been a sort of bluish green. I suspect that the iron(II) sulphate had been stored for a long time and had partly oxidised to iron(III) sulphate. This will react with an alkali, such as NaOH, to give a reddish brown precipitate of hydrated iron(III) oxide (often called iron(III) hydroxide or ferric hydroxide):
6OH-(aq) + 2Fe+++(aq) --> Fe2O3(H2O)n(s)
The 'n' subscript here means that the extent of hydration is variable and indeterminate. Of course, to balance the equation, it would require a 3 at that position.

The gas is hydrogen:
2H2O(l) + Ca(s) --> H2(g) + Ca++(aq) + 2OH-(aq)

You will have got a solution of aluminium chloride and hydrogen gas. (I'm not sure where you are: if you're in the USA, it's spelled 'aluminum'.):
2Al(s) + 6H+(aq) --> 2Al+++(aq) + 3H2(g)

A very similar reaction, but the solution will now contain zinc chloride: Zn(s) + 2H+(aq) --> Zn++(aq) + H2(g)

Hydrochloric acid does not normally react with copper. But, if there were traces of copper(II) oxide on the surface of the metal, they would react, to form a dilute solution of copper(II) chloride:
CuO(s) + 2H+(aq) --> Cu++(aq) + H2O(l) or:
CuO(s) + 2H+(aq) + 4Cl-(aq) --> CuCl4--(aq) + H2O(l)

The zinc will displace the copper, reducing it from the ion Cu++ to the metal and the zince will be oxidised to the ion Zn++. I can't work out why it looked black at the edges: I would have expected it to turn brown - the colour of the copper:
Zn(s) + Cu++(aq) --> Cu(s) + Zn++(aq)

This is a similar displacement reaction. I would expect to see black particles of silver and, when they have settled, a pale blue solution of copper(II) nitrate.
Cu(s) + 2Ag+(aq) --> 2Ag(s) + Cu++(aq)

Because of the limitations of the format, the equations don't look quite right. The plus and minus signs indicating the charges on the ions should be superscripts, i.e. 'up inthe air'. The coefficients of the atoms (as in H2O) should be subscripts. The numbers in brackets indicating the oxidation states (II) and (III) should be in Roman type and the state symbol (I) after H2O is supposed to be a lower-case L. To produce superscripts on Word, highlight the characters and do shift-Ctrl-plus; to produce subscripts, highlight the characters and do Ctrl-equals.

Best of luck,

Allan Deeds.

Answer 4

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