# How would you balance Na2B4O7 + H2SO4 + H2O --> H3BO3 + Na2SO4?

Oct 26, 2015

$N {a}_{2} {B}_{4} {O}_{7}$ + ${H}_{2} S {O}_{4}$ + $5 {H}_{2} O$ $\rightarrow$ $4 {H}_{3} B {O}_{3}$ + $N {a}_{2} S {O}_{4}$

#### Explanation:

You first balance the metals, namely Sodium ($N a$), but in this case it is already balanced, 2 on the left of the arrow and 2 on the right.

Next you do the non-metals, starting with anything but Oxygen ($O$) and Hydrogen ($H$), which in this case is Sulphur but since its part of a polyatomic ion, being Suphate ($S {O}_{4}$) you balance it as a whole, in this case they are both balanced though.

Next is Boron ($B$), on the left there is 4 and on the right there is one, so you add a 4 before the ${H}_{3} B {O}_{3}$ to balance it.

The you balance the Hydrogen ($H$) atoms, on the left there is 4 and on the right there is 12 after putting a 4 in front of (${H}_{3} B {O}_{3}$).
So you add a 5 to the water molecule on the left such that it doesn't effect the Sulphate ion, and in doing so balances the amount of Hydrogens (12 each side).

Everything should now balance, including Oxygen (16 on left and 16 on the right), if it doesn't, just keep trying by putting different values of coefficients before each molecule until it works out.

Remember that the coefficients should be the lowest value possible, and must not be fractions.

Hope I helped :)

Oct 26, 2015

$N {a}_{2} {B}_{4} {O}_{7}$ + ${H}_{2} S {O}_{4}$ +$5 {H}_{2} O$ = $4 {H}_{3} B {O}_{3}$ + $N {a}_{2} S {O}_{4}$

#### Explanation:

It does not matter how long the equation is. You need to always start with tallying the number of atoms (based on the subscripts).

$N {a}_{2} {B}_{4} {O}_{7}$ + ${H}_{2} S {O}_{4}$ +${H}_{2} O$ = ${H}_{3} B {O}_{3}$ + $N {a}_{2} S {O}_{4}$

Left side:

Na = 2
B = 4
O = 7 + 4 + 1 (DO NOT ADD IT UP YET)
H = 2 + 2 (DO NOT ADD IT UP YET)
S= 1

Right side:

Na = 2
B = 1
O = 3 + 4 (DO NOT ADD IT UP YET)
H = 3
S= 1

Start with the element easiest to balance, in this case boron (B). Since B is part of the substance ${H}_{3} B {O}_{3}$, whatever number you choose to multiply with B should also be applied to H and O atoms, respectively.

Left side:

Na = 2
B = 4
O = 7 + 4 + 1
H = 2 + 2
S= 1

Right side:

Na = 2
B = 1 x 4 = 4
O = (3 x 4) + 4
H = (3 x 4) = 12
S= 1

$N {a}_{2} {B}_{4} {O}_{7}$ + ${H}_{2} S {O}_{4}$ +${H}_{2} O$ = $\textcolor{red}{4} {H}_{3} B {O}_{3}$ + $N {a}_{2} S {O}_{4}$

Now, the number of H on your right side is 12. There are two H atoms on your left side that you can choose to multiply by 5 (since both values are 2) to balance. Choose the least complicated substance or the one with least number of other atoms attached to it. In this case, it is ${H}_{2} O$ (2 atoms attached to H from ${H}_{2} O$ vs. 6 atoms attached to H from ${H}_{2} S {O}_{4}$).

Left side:

Na = 2
B = 4
O = 7 + 4 + (1 x 5)
H = 2 + (2 x 5) = 12
S= 1

Right side:

Na = 2
B = 1 x 4 = 4
O = (3 x 4) + 4
H = (3 x 4) = 12
S= 1

$N {a}_{2} {B}_{4} {O}_{7}$ + ${H}_{2} S {O}_{4}$ +$\textcolor{red}{5} {H}_{2} O$ = $\textcolor{red}{4} {H}_{3} B {O}_{3}$ + $N {a}_{2} S {O}_{4}$

Now, look at your tally sheet and see it everything is balanced out.

Left side:

Na = 2
B = 4
O = 7 + 4 + (1 x 5) = 16
H = 2 + (2 x 5) = 12
S= 1

Right side:

Na = 2
B = 1 x 4 = 4
O = (3 x 4) + 4 = 16
H = (3 x 4) = 12
S= 1

The equation is now balanced.

$N {a}_{2} {B}_{4} {O}_{7}$ + ${H}_{2} S {O}_{4}$ +$\textcolor{red}{5} {H}_{2} O$ = $\textcolor{red}{4} {H}_{3} B {O}_{3}$ + $N {a}_{2} S {O}_{4}$