Question

How should this combustion reaction be balanced?
C5Hg + O2 → CO2 + H2O

Answer 1

`\huge \boxed{\mathrm{C_5H_(12)+ 8O_2 \Rightarrow 5CO_2 + 6H_2O}}`

`\rule[225]{225}{2}`

Step-by-step explanation:

`\sf C_5H_(12)+ O_2 \Rightarrow CO_2 + H_2O`

Balancing the Carbon atoms on the right side,

`\sf C_5H_(12)+ O_2 \Rightarrow 5CO_2 + H_2O`

Balancing the Hydrogen atoms on the right side,

`\sf C_5H_(12)+ O_2 \Rightarrow 5CO_2 + 6H_2O`

Balancing the Oxygen atoms on the left side,

`\sf C_5H_(12)+ 8O_2 \Rightarrow 5CO_2 + 6H_2O`

`\rule[225]{225}{2}`

Answer 2

Answer: 4 C5H + 21 O2 > 20 CO2 + 2 H2O

Explanation: there will be 4 moles of C5H, 22 moles of O2 and on the product side 20 moles of CO2 and 2 moles of water.

C atoms is 20 in both reactant and product side

H is 4 atoms each

O have 42 atoms each side.