Chemistry worksheet focusing on converting word equations into balanced chemical equations.

Here are the balanced chemical equations for each word equation. I have worked through them step-by-step to ensure the number of atoms for each element is the same on both sides.

1. zinc chromate + iron III carbonate $\rightarrow$ zinc carbonate + iron III chromate
* Formulas: Zinc is $Zn^{2+}$, Chromate is $CrO_4^{2-}$ ($ZnCrO_4$). Iron(III) is $Fe^{3+}$, Carbonate is $CO_3^{2-}$ ($Fe_2(CO_3)_3$).
* Balancing: We need 3 Zincs and 2 Irons to balance the charges/ions.
* Equation: $3ZnCrO_4 + Fe_2(CO_3)_3 \rightarrow 3ZnCO_3 + Fe_2(CrO_4)_3$

2. barium hypochlorite + iron III carbonate $\rightarrow$ iron III hypochlorite + barium carbonate
* Formulas: Barium is $Ba^{2+}$, Hypochlorite is $ClO^-$ ($Ba(ClO)_2$). Iron(III) is $Fe^{3+}$, Carbonate is $CO_3^{2-}$ ($Fe_2(CO_3)_3$).
* Balancing: We need 3 Bariums and 2 Irons. This creates 6 hypochlorites and 3 carbonates.
* Equation: $3Ba(ClO)_2 + Fe_2(CO_3)_3 \rightarrow 2Fe(ClO)_3 + 3BaCO_3$

3. cobalt II chlorate $\rightarrow$ cobalt II chloride + oxygen
* Formulas: Cobalt(II) is $Co^{2+}$, Chlorate is $ClO_3^-$ ($Co(ClO_3)_2$). Chloride is $Cl^-$ ($CoCl_2$). Oxygen gas is $O_2$.
* Balancing: The reactant has 6 oxygens. To get 6 oxygens on the right, we need 3 $O_2$ molecules.
* Equation: $Co(ClO_3)_2 \rightarrow CoCl_2 + 3O_2$

4. copper I hydroxide + nickel II sulfate $\rightarrow$ nickel II hydroxide + cuprous sulfate
* Formulas: Copper(I) (cuprous) is $Cu^+$, Hydroxide is $OH^-$ ($CuOH$). Nickel(II) is $Ni^{2+}$, Sulfate is $SO_4^{2-}$ ($NiSO_4$). Cuprous sulfate is $Cu_2SO_4$.
* Balancing: We need 2 Copper(I) hydroxides to provide enough copper for the product $Cu_2SO_4$. This also gives us 2 hydroxides for the nickel.
* Equation: $2CuOH + NiSO_4 \rightarrow Ni(OH)_2 + Cu_2SO_4$

5. sulfuric acid + strontium fluoride $\rightarrow$ strontium sulfate + hydrogen fluoride
* Formulas: Sulfuric acid is $H_2SO_4$. Strontium is $Sr^{2+}$, Fluoride is $F^-$ ($SrF_2$). Strontium sulfate is $SrSO_4$. Hydrogen fluoride is $HF$.
* Balancing: There are 2 hydrogens and 2 fluorines in the reactants, so we need 2 $HF$ molecules.
* Equation: $H_2SO_4 + SrF_2 \rightarrow SrSO_4 + 2HF$

6. sodium tartrate + mercury II nitrate $\rightarrow$ sodium nitrate + mercury II tartrate
* Formulas: Sodium is $Na^+$, Tartrate is $C_4H_4O_6^{2-}$ ($Na_2C_4H_4O_6$). Mercury(II) is $Hg^{2+}$, Nitrate is $NO_3^-$ ($Hg(NO_3)_2$).
* Balancing: We have 2 sodiums and 2 nitrates involved in the swap. We need 2 sodium nitrates.
* Equation: $Na_2C_4H_4O_6 + Hg(NO_3)_2 \rightarrow 2NaNO_3 + HgC_4H_4O_6$

7. cadmium + tin IV sulfite $\rightarrow$ tin + cadmium sulfite
* Formulas: Cadmium is $Cd$. Tin(IV) is $Sn^{4+}$, Sulfite is $SO_3^{2-}$ ($Sn(SO_3)_2$). Cadmium is $Cd^{2+}$, so Cadmium sulfite is $CdSO_3$.
* Balancing: There are 2 sulfite groups in the reactant, so we need 2 cadmium sulfites. This requires 2 cadmium atoms.
* Equation: $2Cd + Sn(SO_3)_2 \rightarrow Sn + 2CdSO_3$

8. hydrogen + nitrogen $\rightarrow$ ammonia
* Formulas: Hydrogen gas is $H_2$. Nitrogen gas is $N_2$. Ammonia is $NH_3$.
* Balancing: We need 2 nitrogens on the right ($2NH_3$), which creates 6 hydrogens. We need 3 hydrogen molecules ($3H_2$) on the left to match.
* Equation: $3H_2 + N_2 \rightarrow 2NH_3$

9. lead IV carbonate $\rightarrow$ lead IV oxide + carbon dioxide
* Formulas: Lead(IV) is $Pb^{4+}$, Carbonate is $CO_3^{2-}$ ($Pb(CO_3)_2$). Lead(IV) oxide is $PbO_2$. Carbon dioxide is $CO_2$.
* Balancing: The reactant has 2 carbons and 6 oxygens. The products need to match. We get 1 $PbO_2$ and 2 $CO_2$ (which is 2 carbons and 4 oxygens). Total oxygens on right: $2 + 4 = 6$. It balances perfectly.
* Equation: $Pb(CO_3)_2 \rightarrow PbO_2 + 2CO_2$

10. chromium III hydroxide $\rightarrow$ chromium III oxide + water
* Formulas: Chromium(III) is $Cr^{3+}$, Hydroxide is $OH^-$ ($Cr(OH)_3$). Chromium(III) oxide is $Cr_2O_3$. Water is $H_2O$.
* Balancing: We need 2 chromiums on the left ($2Cr(OH)_3$). This gives 6 hydrogens and 6 oxygens. On the right, $Cr_2O_3$ takes 3 oxygens, leaving 3 oxygens and 6 hydrogens for water. That makes 3 water molecules ($3H_2O$).
* Equation: $2Cr(OH)_3 \rightarrow Cr_2O_3 + 3H_2O$

Final Answer:
1. $3ZnCrO_4 + Fe_2(CO_3)_3 \rightarrow 3ZnCO_3 + Fe_2(CrO_4)_3$
2. $3Ba(ClO)_2 + Fe_2(CO_3)_3 \rightarrow 2Fe(ClO)_3 + 3BaCO_3$
3. $Co(ClO_3)_2 \rightarrow CoCl_2 + 3O_2$
4. $2CuOH + NiSO_4 \rightarrow Ni(OH)_2 + Cu_2SO_4$
5. $H_2SO_4 + SrF_2 \rightarrow SrSO_4 + 2HF$
6. $Na_2C_4H_4O_6 + Hg(NO_3)_2 \rightarrow 2NaNO_3 + HgC_4H_4O_6$
7. $2Cd + Sn(SO_3)_2 \rightarrow Sn + 2CdSO_3$
8. $3H_2 + N_2 \rightarrow 2NH_3$
9. $Pb(CO_3)_2 \rightarrow PbO_2 + 2CO_2$
10. $2Cr(OH)_3 \rightarrow Cr_2O_3 + 3H_2O$