How do you balance ${(N H_{4})}_{2} C O_{3} + C u C l_{2} \to N H_{4} C l + C u C O_{3}$ $(NH_4)_2CO_3 + CuCl_2 -> NH_4Cl + CuCO_3$ ?

Question

How do you balance ${(N H_{4})}_{2} C O_{3} + C u C l_{2} \to N H_{4} C l + C u C O_{3}$ $(NH_4)_2CO_3 + CuCl_2 -> NH_4Cl + CuCO_3$ ?

1 Answer

Impact of this question

1770 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-12T01:15:01

Really, if you examine each molecule and atom as a unit, then almost everything is balanced already.

We see one equivalent of ${Cu}^{2 +}$ $"Cu"^(2+)$ (copper(II)) on each side, and one equivalent of ${CO}_{3}^{2 -}$ $"CO"_3^(2-)$ (carbonate) on each side.

However, we only have one equivalent of ${NH}_{4}^{+}$ $"NH"_4^(+)$ (ammonium) on the right side and one equivalent of ${Cl}^{-}$ $"Cl"^(-)$ (chloride) on the right side, but two equivalents of each on the left side. (The subscripts outside parentheses correspond to the whole polyatomic ion.)

Therefore, all we need is a $2$ $2$ on the ${NH}_{4} Cl$ $color(white)("NH"_4"Cl")$ . As a result, you get (highlight the answer to check):

${({NH}_{4})}_{2} {CO}_{3} + {CuCl}_{2} \to 2 {NH}_{4} Cl + {CuCO}_{3}$ $color(white)(("NH"_4)_2"CO"_3 + "CuCl"_2 -> 2"NH"_4"Cl" + "CuCO"_3)$

Science

Math

Humanities

... and beyond

How do you balance ${(N H_{4})}_{2} C O_{3} + C u C l_{2} \to N H_{4} C l + C u C O_{3}$ $(NH_4)_2CO_3 + CuCl_2 -> NH_4Cl + CuCO_3$ ?

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you balance (NH4)2CO3+CuCl2→NH4Cl+CuCO3(NH_4)_2CO_3 + CuCl_2 -> NH_4Cl + CuCO_3?

1 Answer

Related questions

Impact of this question

How do you balance ${(N H_{4})}_{2} C O_{3} + C u C l_{2} \to N H_{4} C l + C u C O_{3}$ $(NH_4)_2CO_3 + CuCl_2 -> NH_4Cl + CuCO_3$ ?