Respuesta :
To balance an equation, you need to find element that was not spread, in this problem it was Al and Na. You can split the equation into
? naoh → ? na2so4
? al2(so4)3+ → ? al(oh)3
There is two Na on na2so4, so to balance it you need two NaOH
2 naoh → 1 na2so4
? al2(so4)3+ → ? al(oh)3
There is two Al in al2(so4)3 and one Al on al(oh)3, so you need twice amount al(oh)3 to balance the equation.
2 naoh → 1 na2so4
1 al2(so4)3+ → 2 al(oh)3
In the equation above, there is 3 SO4 on reactant but 1 SO4 on the product, so multiply the naoh and na2so4 with 3
6 naoh → 3 na2so4
1 al2(so4)3+ → 2 al(oh)3
The final equation would be:
1 al2(so4)3+ 6 naoh → 2 al(oh)3+ 3 na2so4
The sum of the coefficient would be: 1+6+2+3=12
? naoh → ? na2so4
? al2(so4)3+ → ? al(oh)3
There is two Na on na2so4, so to balance it you need two NaOH
2 naoh → 1 na2so4
? al2(so4)3+ → ? al(oh)3
There is two Al in al2(so4)3 and one Al on al(oh)3, so you need twice amount al(oh)3 to balance the equation.
2 naoh → 1 na2so4
1 al2(so4)3+ → 2 al(oh)3
In the equation above, there is 3 SO4 on reactant but 1 SO4 on the product, so multiply the naoh and na2so4 with 3
6 naoh → 3 na2so4
1 al2(so4)3+ → 2 al(oh)3
The final equation would be:
1 al2(so4)3+ 6 naoh → 2 al(oh)3+ 3 na2so4
The sum of the coefficient would be: 1+6+2+3=12
Answer: 12
Justification:
1) To balance the atoms you can start by Al:
Since there are 2 Al on the left side, add a 2 coefficient in front to Al(OH)₃
That will lead to this transitorye equation:
Al₂(SO₄)₃ + ?NaOH → 2 Al(OH)₃ + ? Na₂SO₄
2) To balance the three SO₄ radicals on the left, add a 3 in front to Na₂SO₄ on the right, leading to:
Al₂(SO₄)₃ + ?NaOH → 2 Al(OH)₃ + 3 Na₂SO₄
3) To balance the six Na atoms on the right, add a 3 in front to NaOH on the left,leading to
Al₂(SO₄)₃ + 6NaOH → 2 Al(OH)₃ + 3 Na₂SO₄
4) Check that all the species are balanced:
Al: 2 on the left and 2 on the right
SO₄: 3 on the left and 3 on the right
Na: 6 on the left and 6 on the right
OH: 6 on the left and 6 on the right.
Then, the equation is balanced.
5) The sum of the coefficients is 1 + 6 = 7 in the left and 2 + 3 = 5 on the right. Total 7 + 5 = 12.
Justification:
1) To balance the atoms you can start by Al:
Since there are 2 Al on the left side, add a 2 coefficient in front to Al(OH)₃
That will lead to this transitorye equation:
Al₂(SO₄)₃ + ?NaOH → 2 Al(OH)₃ + ? Na₂SO₄
2) To balance the three SO₄ radicals on the left, add a 3 in front to Na₂SO₄ on the right, leading to:
Al₂(SO₄)₃ + ?NaOH → 2 Al(OH)₃ + 3 Na₂SO₄
3) To balance the six Na atoms on the right, add a 3 in front to NaOH on the left,leading to
Al₂(SO₄)₃ + 6NaOH → 2 Al(OH)₃ + 3 Na₂SO₄
4) Check that all the species are balanced:
Al: 2 on the left and 2 on the right
SO₄: 3 on the left and 3 on the right
Na: 6 on the left and 6 on the right
OH: 6 on the left and 6 on the right.
Then, the equation is balanced.
5) The sum of the coefficients is 1 + 6 = 7 in the left and 2 + 3 = 5 on the right. Total 7 + 5 = 12.