Total Ionic and Net Ionic Equations Give the total ionic and net ionic equation
ID: 624436 • Letter: T
Question
Total Ionic and Net Ionic Equations Give the total ionic and net ionic equation Mg(NO3)2(aq) + NH3(aq) + Na2HPO4(aq) ------> MgNH4PO4(precipitates) + 2Nano3(aq)Explanation / Answer
To answer this you need to know your solubility rules. Important rules for this problem: All Group I element compounds are soluble. All NO3 compounds are soluble. All CO3 compounds are INSOLUBLE (except for group I elements & NH3) So your reaction is: Mg(NO3)2(aq) + Na2CO3(aq) --> MgCO3(s) + NaNO3(aq) Aqueous compounds will dissociate in water giving you: Mg2+(aq) + NO3-(aq) + Na+(aq) + CO3 2-(aq) --> MgCO3(s) + Na+(aq) + NO3-(aq) Now you cross out anything you have on both sides. These are your spectator ions. After that your net ionic equation will give you.... Mg2+(aq) + CO3 2-(aq) --> MgCO3(s) You need to balance the molecular equation first Mg(NO3)2 + Na2CO3 --> MgCO3 + 2NaNO3 Find the total ionic equation, which is obtained by breaking the soluble molecules into their respective ions (one positive and one negative). Insoluble molecules can't be broken. You need to know about the solubility rules in order to write the equation. See here http://www.ausetute.com.au/solrules.html Mg²? + 2NO3? + 2Na? + CO3²? --> MgCO3 + 2Na? + 2NO3? MgCO3 is insoluble, so it can't be broken apart To write the net ionic equation, cancel out the ions that appear on both sides, which are Na? and NO3?, you get Mg²? + CO3²? --> MgCO3