Question

add the correct coefficients to balance the following skeleton equation: _h₂co₃→_h₂o+_co₂ be careful with the oxygens! _fe₃o₄+_h₂→_fe+___h₂o

Explanation:

Step1: Balance the first equation \(H_{2}CO_{3}\to H_{2}O + CO_{2}\)

• Count atoms:
• Reactants (\(H_{2}CO_{3}\)): \(H = 2\), \(C=1\), \(O = 3\)
• Products (\(H_{2}O+CO_{2}\)): \(H = 2\), \(C = 1\), \(O=1 + 2=3\)
• Check balance:
• For \(H\): \(2 = 2\)
• For \(C\): \(1=1\)
• For \(O\): \(3=3\)
• So the balanced equation is \(1H_{2}CO_{3}\to1H_{2}O + 1CO_{2}\)

Step2: Balance the second equation \(Fe_{3}O_{4}+H_{2}\to Fe + H_{2}O\)

• Balance Fe:
• Reactant (\(Fe_{3}O_{4}\)) has \(Fe = 3\). So put \(3\) in front of \(Fe\) in products: \(Fe_{3}O_{4}+H_{2}\to3Fe + H_{2}O\)
• Balance O:
• Reactant (\(Fe_{3}O_{4}\)) has \(O = 4\). So put \(4\) in front of \(H_{2}O\) in products: \(Fe_{3}O_{4}+H_{2}\to3Fe + 4H_{2}O\)
• Balance H:
• Products (\(4H_{2}O\)) has \(H=8\). So put \(4\) in front of \(H_{2}\) in reactants: \(Fe_{3}O_{4}+4H_{2}\to3Fe + 4H_{2}O\)

Answer:

• First equation: \(1H_{2}CO_{3}\to1H_{2}O + 1CO_{2}\)
• Second equation: \(1Fe_{3}O_{4}+4H_{2}\to3Fe + 4H_{2}O\)