Chemistry, 27.03.2020 04:07 itsme123427

Indicate the number corresponding to each coefficient required to write the balanced equation for the
reaction:
aH2SO4 + bKOH → cH2O+ dK2S04
a=
b=
c=
d=

Answers: 3

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Answer from: ErrorNameTaken505

$\rm H_2SO_4 + 2\, KOH \to 2\, H_2O + K_2SO_4$ .

Explanation:

Indeed it is possible to solve this problem with system of multiple equations. This explanation will give an approach that gives the same result with less calculations.

Start by finding the most complex species in the equation and setting its coefficient to . That species should as many elements as possible. For example, in this question, $\rm H_2SO_4$ , $\rm K_2SO_4$ , and $\rm KOH$ all contain three elements each. Nonetheless, $\rm H_2SO_4$ and $\rm K_2SO_4$ are more preferable for they contain more atoms.

For example, in case $\rm H_2SO_4$ is chosen as the most complex species. Set its coefficient to :

$1 \, \mathrm{H_2SO_4} + b\, \mathrm{KOH} \to c\, \mathrm{H_2O} + d\, \mathrm{K_2SO_4}$ .

There are two $\rm H$ atoms, one $\rm S$ atom, and four $\rm O$ atoms in that $\rm H_2SO_4$ . Since $\rm KOH$ also contains $\rm H$ and $\rm O$ , it is not certain how many $\rm H$ and $\rm O$ on the left-hand side in total. However, on the left-hand side, $\rm S$ appears only in $\rm H_2SO_4$ . Therefore, it is certain that there is only one $\rm S$ in the left-hand side of the equation.

In chemical reaction, atoms of an element are neither created nor destroyed. Therefore, the number of $\rm S$ atoms on the right-hand side should be the same as that on the left-hand side. Therefore, there should also be only one $\rm S$ atom on the right-hand side.

On the right-hand side, $\rm K_2SO_4$ is the only compound that contains $\rm S$ . Besides, each formula unit of $\rm K_2SO_4$ contains exactly one $\rm S$ . The only way to get exactly one $\rm S$ atom on the right-hand side is to set the coefficient of $\rm K_2SO_4$ to one, as well. The equation will then look like:

$1 \, \mathrm{H_2SO_4} + b\, \mathrm{KOH} \to c\, \mathrm{H_2O} + 1\, \mathrm{K_2SO_4}$ .

With a similar logic, on the right-hand side, $\rm K_2SO_4$ is the only species with $\rm K$ $\rm K$ atoms. There would be exactly two $\rm K$ atoms on each side of the equation. On the left-hand side, $\rm KOH$ is the only species with $\rm K$ atoms. Each formula unit of $\rm KOH$ contains one $\rm K$ atom. Therefore, the coefficient of $\rm KOH$ should be 2/1 = 2 .

$1 \, \mathrm{H_2SO_4} + 2\, \mathrm{KOH} \to c\, \mathrm{H_2O} + 1\, \mathrm{K_2SO_4}$ .

It is now certain that there are $1 \times 4 + 2 \time 1 = 6$ (six) $\rm O$ atoms on the left-hand side of the equation. With a similar logic, there should also be six $\rm O$ atoms on the right-hand side of the equation. Four of these would come from $\rm K_2SO_4$ . The other two would come from $\rm H_2O$ .

Each $\rm H_2O$ molecule contains one $\rm O$ atom. As a result, there needs to be 2 / 1 = 2 (two) $\rm H_2O$ molecules on the left-hand side.

Hence the equation:

$1 \, \mathrm{H_2SO_4} + 2\, \mathrm{KOH} \to 2\, \mathrm{H_2O} + 1\, \mathrm{K_2SO_4}$ .

Double-check the work by making sure that the number of $\rm H$ atoms is also the same on both sides of the equation.

Answer from: Quest

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