Question

write the formulas of the compounds produced from the listed ions.

Explanation:

Step1: Combine ions

When combining ions to form compounds, the overall charge of the compound must be zero. We use the criss - cross method where the magnitude of the charge of one ion becomes the sub - script of the other ion.

Step2: For $Na^+$ and $Cl^-$

The compound is $NaCl$ since the + 1 charge of $Na^+$ and - 1 charge of $Cl^-$ balance each other.

Step3: For $Na^+$ and $CO_3^{2 - }$

Using the criss - cross method, we get $Na_2CO_3$ as we need 2 $Na^+$ ions to balance the - 2 charge of $CO_3^{2 - }$.

Step4: For $Na^+$ and $OH^-$

The compound is $NaOH$.

Step5: For $Na^+$ and $SO_4^{2 - }$

We get $Na_2SO_4$.

Step6: For $Na^+$ and $PO_4^{3 - }$

We get $Na_3PO_4$.

Step7: For $Na^+$ and $NO_3^-$

The compound is $NaNO_3$.

Step8: For $NH_4^+$ and $Cl^-$

The compound is $NH_4Cl$.

Step9: For $NH_4^+$ and $CO_3^{2 - }$

We get $(NH_4)_2CO_3$.

Step10: For $NH_4^+$ and $OH^-$

The compound is $NH_4OH$.

Step11: For $NH_4^+$ and $SO_4^{2 - }$

We get $(NH_4)_2SO_4$.

Step12: For $NH_4^+$ and $PO_4^{3 - }$

We get $(NH_4)_3PO_4$.

Step13: For $NH_4^+$ and $NO_3^-$

The compound is $NH_4NO_3$.

Step14: For $K^+$ and $Cl^-$

The compound is $KCl$.

Step15: For $K^+$ and $CO_3^{2 - }$

We get $K_2CO_3$.

Step16: For $K^+$ and $OH^-$

The compound is $KOH$.

Step17: For $K^+$ and $SO_4^{2 - }$

We get $K_2SO_4$.

Step18: For $K^+$ and $PO_4^{3 - }$

We get $K_3PO_4$.

Step19: For $K^+$ and $NO_3^-$

The compound is $KNO_3$.

Step20: For $Ca^{2 + }$ and $Cl^-$

We get $CaCl_2$.

Step21: For $Ca^{2 + }$ and $CO_3^{2 - }$

The compound is $CaCO_3$.

Step22: For $Ca^{2 + }$ and $OH^-$

We get $Ca(OH)_2$.

Step23: For $Ca^{2 + }$ and $SO_4^{2 - }$

The compound is $CaSO_4$.

Step24: For $Ca^{2 + }$ and $PO_4^{3 - }$

We get $Ca_3(PO_4)_2$.

Step25: For $Ca^{2 + }$ and $NO_3^-$

We get $Ca(NO_3)_2$.

Step26: For $Mg^{2 + }$ and $Cl^-$

We get $MgCl_2$.

Step27: For $Mg^{2 + }$ and $CO_3^{2 - }$

The compound is $MgCO_3$.

Step28: For $Mg^{2 + }$ and $OH^-$

We get $Mg(OH)_2$.

Step29: For $Mg^{2 + }$ and $SO_4^{2 - }$

The compound is $MgSO_4$.

Step30: For $Mg^{2 + }$ and $PO_4^{3 - }$

We get $Mg_3(PO_4)_2$.

Step31: For $Mg^{2 + }$ and $NO_3^-$

We get $Mg(NO_3)_2$.

Step32: For $Zn^{2 + }$ and $Cl^-$

We get $ZnCl_2$.

Step33: For $Zn^{2 + }$ and $CO_3^{2 - }$

The compound is $ZnCO_3$.

Step34: For $Zn^{2 + }$ and $OH^-$

We get $Zn(OH)_2$.

Step35: For $Zn^{2 + }$ and $SO_4^{2 - }$

The compound is $ZnSO_4$.

Step36: For $Zn^{2 + }$ and $PO_4^{3 - }$

We get $Zn_3(PO_4)_2$.

Step37: For $Zn^{2 + }$ and $NO_3^-$

We get $Zn(NO_3)_2$.

Step38: For $Fe^{3 + }$ and $Cl^-$

We get $FeCl_3$.

Step39: For $Fe^{3 + }$ and $CO_3^{2 - }$

We get $Fe_2(CO_3)_3$.

Step40: For $Fe^{3 + }$ and $OH^-$

We get $Fe(OH)_3$.

Step41: For $Fe^{3 + }$ and $SO_4^{2 - }$

We get $Fe_2(SO_4)_3$.

Step42: For $Fe^{3 + }$ and $PO_4^{3 - }$

The compound is $FePO_4$.

Step43: For $Fe^{3 + }$ and $NO_3^-$

We get $Fe(NO_3)_3$.

Step44: For $Al^{3 + }$ and $Cl^-$

We get $AlCl_3$.

Step45: For $Al^{3 + }$ and $CO_3^{2 - }$

We get $Al_2(CO_3)_3$.

Step46: For $Al^{3 + }$ and $OH^-$

We get $Al(OH)_3$.

Step47: For $Al^{3 + }$ and $SO_4^{2 - }$

We get $Al_2(SO_4)_3$.

Step48: For $Al^{3 + }$ and $PO_4^{3 - }$

The compound is $AlPO_4$.

Step49: For $Al^{3 + }$ and $NO_3^-$

We get $Al(NO_3)_3$.

Step50: For $Co^{3 + }$ and $Cl^-$

We get $CoCl_3$.

Step51: For $Co^{3 + }$ and $CO_3^{2 - }$

We get $Co_2(CO_3)_3$.

Step52: For $Co^{3 + }$ and $OH^-$

We get $Co(OH)_3$.

Step53: For $Co^{3 + }$ and $SO_4^{2 - }$

We get $Co_2(SO_4)_3$.

Step54: For $Co^{3 + }$ and $PO_4^{3 - }$

The compound is $CoPO_4$.

Step55: For $Co^{3 + }$ and $NO_3^-$

We get $Co(NO_3)_3$.

Step56: For $Fe^{2 + }$ and $Cl^-$

We get $FeCl_2$.

Step57: For $Fe^{2 + }$ and $CO_3^{2 - }$

The compound is $FeCO_3$.

Step58: For $Fe^{2 + }$ and $OH^-$

We get $Fe(OH)_2$.

Step59: For $Fe^{2 + }$ and $SO_4^{2 - }$

The compound is $FeSO_4$.

Step60: For $Fe^{2 + }$ and $PO_4^{3 - }$

We get $Fe_3(PO_4)_2$.

Step61: For $Fe^{2 + }$ and $NO_3^-$

We get $Fe(NO_3)_2$.

Step62: For $H^+$ and $Cl^-$

The compound is $HCl$.

Step63: For $H^+$ and $CO_3^{2 - }$

We get $H_2CO_3$.

Step64: For $H^+$ and $OH^-$

The compound is $H_2O$.

Step65: For $H^+$ and $SO_4^{2 - }$

We get $H_2SO_4$.

Step66: For $H^+$ and $PO_4^{3 - }$

We get $H_3PO_4$.

Step67: For $H^+$ and $NO_3^-$

The compound is $HNO_3$.

Answer:

	$Cl^-$	$CO_3^{2 - }$	$OH^-$	$SO_4^{2 - }$	$PO_4^{3 - }$	$NO_3^-$
$NH_4^+$	$NH_4Cl$	$(NH_4)_2CO_3$	$NH_4OH$	$(NH_4)_2SO_4$	$(NH_4)_3PO_4$	$NH_4NO_3$
$K^+$	$KCl$	$K_2CO_3$	$KOH$	$K_2SO_4$	$K_3PO_4$	$KNO_3$
$Ca^{2 + }$	$CaCl_2$	$CaCO_3$	$Ca(OH)_2$	$CaSO_4$	$Ca_3(PO_4)_2$	$Ca(NO_3)_2$
$Mg^{2 + }$	$MgCl_2$	$MgCO_3$	$Mg(OH)_2$	$MgSO_4$	$Mg_3(PO_4)_2$	$Mg(NO_3)_2$
$Zn^{2 + }$	$ZnCl_2$	$ZnCO_3$	$Zn(OH)_2$	$ZnSO_4$	$Zn_3(PO_4)_2$	$Zn(NO_3)_2$
$Fe^{3 + }$	$FeCl_3$	$Fe_2(CO_3)_3$	$Fe(OH)_3$	$Fe_2(SO_4)_3$	$FePO_4$	$Fe(NO_3)_3$
$Al^{3 + }$	$AlCl_3$	$Al_2(CO_3)_3$	$Al(OH)_3$	$Al_2(SO_4)_3$	$AlPO_4$	$Al(NO_3)_3$
$Co^{3 + }$	$CoCl_3$	$Co_2(CO_3)_3$	$Co(OH)_3$	$Co_2(SO_4)_3$	$CoPO_4$	$Co(NO_3)_3$
$Fe^{2 + }$	$FeCl_2$	$FeCO_3$	$Fe(OH)_2$	$FeSO_4$	$Fe_3(PO_4)_2$	$Fe(NO_3)_2$
$H^+$	$HCl$	$H_2CO_3$	$H_2O$	$H_2SO_4$	$H_3PO_4$	$HNO_3$