Question

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Explanation:

Step1: Balance Fe atoms

For the equation \(Fe + H_{2}SO_{4}\to Fe_{2}(SO_{4})_{3}+H_{2}\), in \(Fe_{2}(SO_{4})_{3}\) there are 2 Fe atoms. So we put 2 in front of Fe: \(2Fe + H_{2}SO_{4}\to Fe_{2}(SO_{4})_{3}+H_{2}\)

Step2: Balance \(SO_{4}\) groups

There are 3 \(SO_{4}\) groups in \(Fe_{2}(SO_{4})_{3}\). So we put 3 in front of \(H_{2}SO_{4}\): \(2Fe + 3H_{2}SO_{4}\to Fe_{2}(SO_{4})_{3}+H_{2}\)

Step3: Balance H atoms

On the left - hand side, from \(3H_{2}SO_{4}\), there are \(3\times2 = 6\) H atoms. So we put 3 in front of \(H_{2}\): \(2Fe + 3H_{2}SO_{4}=Fe_{2}(SO_{4})_{3}+3H_{2}\)

Step4: For \(C_{2}H_{6}+O_{2}\to H_{2}O + CO_{2}\)

First, balance C atoms. There are 2 C atoms in \(C_{2}H_{6}\), so put 2 in front of \(CO_{2}\): \(C_{2}H_{6}+O_{2}\to H_{2}O + 2CO_{2}\)

Step5: Balance H atoms

There are 6 H atoms in \(C_{2}H_{6}\), so put 3 in front of \(H_{2}O\): \(C_{2}H_{6}+O_{2}\to 3H_{2}O + 2CO_{2}\)

Step6: Balance O atoms

On the right - hand side, there are \(3 + 4=7\) O atoms. So put \(\frac{7}{2}\) in front of \(O_{2}\). Multiply through by 2 to get whole numbers: \(2C_{2}H_{6}+7O_{2}=6H_{2}O + 4CO_{2}\)

Step7: For \(KOH + H_{3}PO_{4}\to K_{3}PO_{4}+H_{2}O\)

Balance K atoms. There are 3 K atoms in \(K_{3}PO_{4}\), so put 3 in front of \(KOH\): \(3KOH + H_{3}PO_{4}\to K_{3}PO_{4}+H_{2}O\)

Step8: Balance H and O atoms

On the left - hand side, there are \(3 + 3=6\) H atoms and \(3 + 4 = 7\) O atoms. On the right - hand side, in \(K_{3}PO_{4}\) there are 4 O atoms. From \(H_{2}O\), if we put 3 in front of \(H_{2}O\) (since \(3\times2=6\) H atoms and \(3\times1 = 3\) O atoms), the equation becomes \(3KOH + H_{3}PO_{4}=K_{3}PO_{4}+3H_{2}O\)

Step9: For \(SnO_{2}+H_{2}\to Sn + H_{2}O\)

Balance O atoms. There are 2 O atoms in \(SnO_{2}\), so put 2 in front of \(H_{2}O\): \(SnO_{2}+H_{2}\to Sn + 2H_{2}O\)

Step10: Balance H atoms

There are 4 H atoms on the right - hand side (from \(2H_{2}O\)), so put 2 in front of \(H_{2}\): \(SnO_{2}+2H_{2}=Sn + 2H_{2}O\)

Step11: For \(NH_{3}+O_{2}\to NO + H_{2}O\)

Use the oxidation - reduction method or trial - and - error.
First, balance N atoms (already 1 on each side).
Balance H atoms: There are 3 H atoms in \(NH_{3}\), so put \(\frac{3}{2}\) in front of \(H_{2}O\). Balance O atoms: Let's rewrite the equation as \(NH_{3}+\frac{5}{4}O_{2}\to NO+\frac{3}{2}H_{2}O\). Multiply through by 4 to get whole numbers: \(4NH_{3}+5O_{2}=4NO + 6H_{2}O\)

Answer:

1. \(2Fe + 3H_{2}SO_{4}=Fe_{2}(SO_{4})_{3}+3H_{2}\)
2. \(2C_{2}H_{6}+7O_{2}=6H_{2}O + 4CO_{2}\)
3. \(3KOH + H_{3}PO_{4}=K_{3}PO_{4}+3H_{2}O\)
4. \(SnO_{2}+2H_{2}=Sn + 2H_{2}O\)
5. \(4NH_{3}+5O_{2}=4NO + 6H_{2}O\)