balance molecular formula, total ionic formula and net ionic formula of Sr(NO3)2 + Al2(SO4)3. Thank you
To determine the balanced molecular formula, total ionic formula, and net ionic formula of the reaction between strontium nitrate (Sr(NO3)2) and aluminum sulfate (Al2(SO4)3), we need to first write the individual formulas and then balance the equation.
1. Write the formulas of the reactants:
- Strontium nitrate: Sr(NO3)2
- Aluminum sulfate: Al2(SO4)3
2. Write the balanced equation:
3Sr(NO3)2 + 2Al2(SO4)3 -> 3SrSO4 + 2Al(NO3)3
3. Balance the equation:
To balance the equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation.
- Strontium (Sr): We have 3 Sr atoms on the left side, so we need 3 Sr atoms on the right side by adding a coefficient of 3 in front of SrSO4.
- Nitrogen (N): We have 6 NO3 groups on the left side (2 NO3 groups per Sr(NO3)2), so we need 6 NO3 groups on the right side by adding a coefficient of 3 in front of Al(NO3)3.
- Oxygen (O): We have 24 oxygen atoms on the left side (12 from NO3 and 12 from SO4 groups). On the right side, we have 12 oxygen atoms from SrSO4. To balance the oxygen, we add 12 more oxygen atoms by multiplying the coefficient of Al(NO3)3 by 6.
The balanced equation becomes:
3Sr(NO3)2 + 2Al2(SO4)3 -> 3SrSO4 + 6Al(NO3)3
4. Determine the ionic compounds and their charges:
- Strontium nitrate: Sr2+, NO3-
- Aluminum sulfate: Al3+, SO42-
5. Write the total ionic formula:
The total ionic formula shows all the ions present in the reaction, including spectator ions (ions that appear on both sides of the equation without undergoing any changes).
Total ionic formula:
3Sr2+ + 6NO3- + 2Al3+ + 3SO42- -> 3SrSO4 + 6Al3+ + 6NO3-
6. Write the net ionic formula:
The net ionic formula shows only the ions that actively participate in the reaction, excluding spectator ions.
Net ionic formula:
3Sr2+ + 3SO42- -> 3SrSO4
Please note that the charges of the ions in the formulas may vary depending on the context.
To balance the molecular formula of Sr(NO3)2 + Al2(SO4)3, we need to make sure that the total number of each element is equal on both sides.
The molecular formula of strontium nitrate is Sr(NO3)2, which means that it contains one strontium atom (Sr), two nitrate ions (NO3-).
Similarly, the molecular formula of aluminum sulfate is Al2(SO4)3, indicating that it consists of two aluminum atoms (Al) and three sulfate ions (SO4^2-).
To balance the equation, we need to ensure that the number of each element is the same on both sides.
Balanced molecular equation:
Sr(NO3)2 + Al2(SO4)3 -> SrSO4 + 2Al(NO3)3
Now, let's move on to the total ionic equation:
In the total ionic equation, we break down soluble compounds into their constituent ions.
Sr(NO3)2 (aq) + Al2(SO4)3 (aq) -> SrSO4 (s) + 2Al(NO3)3 (aq)
Note that (aq) indicates ions dissolved in water, and (s) denotes a solid precipitate formed.
Now, let's move on to the net ionic equation:
The net ionic equation eliminates spectator ions (which appear on both sides of the equation) and only includes the species directly involved in the reaction.
In this case, the spectator ions are Sr^2+ and NO3^-, as they appear on both sides.
Net ionic equation:
2Al^3+ (aq) + 3SO4^2- (aq) -> Al2(SO4)3 (s)
So, the balanced molecular formula, total ionic formula, and net ionic formula of Sr(NO3)2 + Al2(SO4)3 are as follows:
Molecular formula: Sr(NO3)2 + Al2(SO4)3 -> SrSO4 + 2Al(NO3)3
Total ionic formula: Sr^2+ (aq) + 2NO3^- (aq) + 2Al^3+ (aq) + 3SO4^2- (aq) -> SrSO4 (s) + 2Al(NO3)3 (aq)
Net ionic formula: 2Al^3+ (aq) + 3SO4^2- (aq) -> Al2(SO4)3 (s)
3Sr(NO3)2(aq) + Al2(SO4)3(aq) ==> 3SrSO4(s) + 2Al(NO3)3
Split positive from negative ions for the total iionic formula. You can do that, start by
3Sr^+2(aq) + 6NO3^-(aq) + 2Al^+3(aq) + 3SO4^-2(aq)==> etc.
net ionic equation.
3Sr^+2(aq) + 3SO4^-2(aq) ==> 3SrSO4(s)