Question

balance the following equation: kmno₄ + hcl → kcl + mncl₂ + cl₂ + h₂o your answer balance the following equation: nh₃ + o₂ → no + h₂o your answer balance the following equation: c₃h₈o + o₂ → co₂ + h₂o your answer balance the following equation: fes₂ + o₂ → fe₂o₃ + so₂

Explanation:

Step1: Balance the first equation \(KMnO_{4}+HCl

ightarrow KCl + MnCl_{2}+Cl_{2}+H_{2}O\)

• Assign oxidation states:
• In \(KMnO_{4}\), \(Mn\) is in \(+7\) state. In \(MnCl_{2}\), \(Mn\) is in \(+2\) state. So \(Mn\) gains \(5e^{-}\) per atom.
• In \(HCl\), \(Cl\) is in \(-1\) state. In \(Cl_{2}\), \(Cl\) is in \(0\) state. So \(Cl\) loses \(1e^{-}\) per atom.
• To balance electrons, multiply \(KMnO_{4}\) by \(2\) (total \(10e^{-}\) gain) and \(Cl_{2}\) by \(5\) (total \(10e^{-}\) loss).
• Now the equation becomes \(2KMnO_{4}+HCl

ightarrow 2KCl + 2MnCl_{2}+5Cl_{2}+H_{2}O\)

• Balance \(K\): already balanced.
• Balance \(Mn\): already balanced.
• Balance \(Cl\): On left, \(HCl\) provides \(Cl\). On right, \(2KCl + 2MnCl_{2}+5Cl_{2}\) has \(2 + 4+ 10=16Cl\). So \(HCl\) is \(16\).
• Balance \(H\) and \(O\): With \(16HCl\), we have \(16H\). On right, \(H_{2}O\) has \(8H_{2}O\) (from \(16H\)). And \(O\) from \(2KMnO_{4}\) is \(8O\), which matches \(8H_{2}O\).
• Final balanced equation: \(2KMnO_{4}+16HCl = 2KCl + 2MnCl_{2}+5Cl_{2}+8H_{2}O\)

Step2: Balance the second equation \(NH_{3}+O_{2}

ightarrow NO + H_{2}O\)

• Use the algebraic method. Let coefficients be \(aNH_{3}+bO_{2}=cNO + dH_{2}O\)
• \(N\): \(a = c\)
• \(H\): \(3a=2d\)
• \(O\): \(2b=c + d\)
• Let \(a = 4\), then \(c = 4\), \(d = 6\) (from \(H\) balance). Substitute into \(O\) balance: \(2b=4 + 6\Rightarrow b = 5\)
• Final balanced equation: \(4NH_{3}+5O_{2}=4NO + 6H_{2}O\)

Step3: Balance the third equation \(C_{3}H_{8}O+O_{2}

ightarrow CO_{2}+H_{2}O\)

• Balance \(C\): \(C_{3}H_{8}O\) has \(3C\), so \(CO_{2}\) is \(3\)
• Balance \(H\): \(C_{3}H_{8}O\) has \(8H\), so \(H_{2}O\) is \(4\)
• Now equation: \(C_{3}H_{8}O+O_{2}

ightarrow 3CO_{2}+4H_{2}O\)

• Balance \(O\): Left has \(1\) (from \(C_{3}H_{8}O\)) + \(2b\) (from \(O_{2}\)). Right has \(6 + 4=10\). So \(1+2b = 10\Rightarrow b=\frac{9}{2}\). Multiply all by \(2\) to get rid of fraction.
• Final balanced equation: \(2C_{3}H_{8}O+9O_{2}=6CO_{2}+8H_{2}O\)

Step4: Balance the fourth equation \(FeS_{2}+O_{2}

ightarrow Fe_{2}O_{3}+SO_{2}\)

• Balance \(Fe\): \(2FeS_{2}+O_{2}

ightarrow Fe_{2}O_{3}+SO_{2}\)

• Balance \(S\): \(2FeS_{2}+O_{2}

ightarrow Fe_{2}O_{3}+4SO_{2}\)

• Balance \(O\): Right has \(3 + 8 = 11O\). Left has \(2bO\). Let \(b=\frac{11}{2}\). Multiply all by \(2\)
• Final balanced equation: \(4FeS_{2}+11O_{2}=2Fe_{2}O_{3}+8SO_{2}\)

Answer:

1. \(2KMnO_{4}+16HCl = 2KCl + 2MnCl_{2}+5Cl_{2}+8H_{2}O\)
2. \(4NH_{3}+5O_{2}=4NO + 6H_{2}O\)
3. \(2C_{3}H_{8}O+9O_{2}=6CO_{2}+8H_{2}O\)
4. \(4FeS_{2}+11O_{2}=2Fe_{2}O_{3}+8SO_{2}\)