`~Ca(OH)_2 + ~(NH_4)_2SO_4 -gt ~CaSO_4 + ~NH_3 + ~H_2O`
Step 1: Make a list of the number of each type of atom on the reactant and product sides of the equation.
`~Ca(OH)_2 + ~(NH_4)_2SO_4 -gt ~CaSO_4 + ~NH_3 + ~H_2O`
Reactants Products
Ca: 1 Ca: 1
O: 6 O: 5
H: 10 H: 5
N: 2 N: 1
S: 1 S: 1
Step 2: Pick an unbalanced atom to start with. Add a coefficient to balance the unbalanced atom.
Let’s start with N. Ask yourself “What number can I multiply with the subscript of N on the product side of the equation that will give me the same number of N atoms as on the reactant side of the equation?” The answer would be 2.
So, place a coefficient of 2 in front of the compound on the product side of the equation that contains N. This changes the number of N atoms on the product side to 2 AND changes the number of H atoms to 8.
`~Ca(OH)_2 + ~(NH_4)_2SO_4 -gt ~CaSO_4 + ~2NH_3 + ~H_2O`
Reactants Products
Ca: 1 Ca: 1
O: 6 O: 5
H: 10 H: 8
N: 2 N: 2
S: 1 S: 1
Step 3: Choose another unbalanced atom in the equation. Add a coefficient to balance the unbalanced atom.
Let’s balance H next. Notice that H is present in two compounds on the product side of the equation. Ask yourself “What number can I multiply with the subscript of H in one of the H-containing compounds on the product side that, when added to the other H-containing compound, will give me the same number of H atoms as on the reactant side of the equation?"
If we place a coefficient of 2 in front of the compound `~H_2O` , the total number of H atoms on the product side will equal the number of H atoms on the reactant side. Placing a coefficient of 2 in front of the compound `~H_2O` also changes the total number of O atoms on the product side to 6.
`~Ca(OH)_2 + ~(NH_4)_2SO_4 -gt ~CaSO_4 + ~2NH_3 + ~2H_2O`
Reactants Products
Ca: 1 Ca: 1
O: 6 O: 6
H: 10 H: 10
N: 2 N: 2
S: 1 S: 1
The reaction is now balanced.
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