Question

lowing double displacement reaction equations (assume all reac- and balance the following cumplete and as go to products). Agel & Na NO OB NaCl + AgNO₃ - BaCl2 + H2SO, — NaOH + HCl - Na,Cog + HCI HgSO4 + NH4OH FeCl, + NH,OH — Na2SO3 + HCI — K,Cro, + Pb(NO3), — NaC,H,O2 + HCl — NaOH + NH.NO пояс BiClg + H2S 12. K.C,0+ HCI — HAPO, + Ca(OH)2 - HNO (NH) CO + NiBr, K.CO2 +

Answer 1

In balancing a chemical reaction, we have to make each of the atoms present in the reactant (i.e on the left side of the equation) equal in number to the product (i.e on the right side of the equation)

For this we have to multiply the required number before the written molecular formula of the compound.

Now in a double displacement recation, we react two salts. Here the ions of the salts get interchanged to form two new salts in prodcut.

So in this way-

1-

NaCl + AgNO3 ---------->

(Na+ + Cl-) + (Ag+ + NO3-) -------->

(Na+ + NO3-) + (Ag+ + Cl-) -------> NaNO3 + AgCl

2- BaCl2 + H2SO4 ---------->

(Ba+2 + 2Cl-) + (2H+ + SO42-) -------->

(Ba+2 + SO42-) + (2H+ + 2Cl-) -------> BaSO4 + 2HCl

3-

NaOH + HCl ---------->

(Na+ + OH-) + (H+ + Cl-) -------->

(Na+ + Cl-) + (H+ + OH-) -------> NaCl + H2O

4-

Na2CO3 + 2HCl ---------->

(2Na+ + CO3-2) + (2H+ + 2Cl-) -------->

(2Na+ + 2Cl-​​​​​​​) + (2H+ + CO3-2) -------> 2NaCl + H2CO3

5-

H2SO4 + 2NH4OH ---------->

(2H+ + SO42-) + 2(NH4+ + OH-) -------->

(2H+ + 2OH-​​​​​​​) + (2NH4+ + SO42-) -------> 2H2O + (NH4)2SO4

6-

FeCl3 + 3NH4OH ---------->

(Fe+3 + 3Cl-) + 3(NH4+ + OH-) -------->

(Fe+3 + 3OH-​​​​​​​) + (3NH4+ + 3Cl-) -------> Fe(OH)3 + 3NH4Cl

Similarly the rest will be done