Question

Question was provided via image upload.

Explanation:

Step1: Balance the first equation

For \(Fe + H_2SO_4
ightarrow Fe_2(SO_4)_3 + H_2\)

• Balance \(Fe\) atoms: There are 2 \(Fe\) atoms in \(Fe_2(SO_4)_3\), so put 2 before \(Fe\)
• Balance \(SO_4\) groups: There are 3 \(SO_4\) groups in \(Fe_2(SO_4)_3\), so put 3 before \(H_2SO_4\)
• Balance \(H\) atoms: There are 6 \(H\) atoms in \(3H_2SO_4\), so put 3 before \(H_2\)

Step2: Balance the second equation

For \(C_2H_6+O_2
ightarrow H_2O + CO_2\)

• Balance \(C\) atoms: There are 2 \(C\) atoms in \(C_2H_6\), so put 2 before \(CO_2\)
• Balance \(H\) atoms: There are 6 \(H\) atoms in \(C_2H_6\), so put 3 before \(H_2O\)
• Balance \(O\) atoms: There are \(2\times2 + 3\times1=7\) \(O\) atoms on the right - hand side, so put \(\frac{7}{2}\) before \(O_2\). But to get whole - number coefficients, multiply all coefficients by 2. So we have \(2C_2H_6+7O_2

ightarrow6H_2O + 4CO_2\)

Step3: Balance the third equation

For \(KOH + H_3PO_4
ightarrow K_3PO_4+H_2O\)

• Balance \(K\) atoms: There are 3 \(K\) atoms in \(K_3PO_4\), so put 3 before \(KOH\)
• Balance \(H\) atoms: There are \(3 + 3=6\) \(H\) atoms on the left - hand side (from \(3KOH\) and \(H_3PO_4\)), so put 3 before \(H_2O\)
• Balance \(PO_4\) groups: There is 1 \(PO_4\) group on both sides

Answer:

1. \(2Fe+3H_2SO_4

ightarrow Fe_2(SO_4)_3 + 3H_2\)

1. \(2C_2H_6+7O_2

ightarrow6H_2O + 4CO_2\)

1. \(3KOH + H_3PO_4

ightarrow K_3PO_4+3H_2O\)