Question

Determine the mass of carbon iv oxide ,produced on burning 104g of ethyne​

Answer 1

163 grams (3 sig. fig.).

Step-by-step explanation

• Formula of carbon(IV) oxide (a.k.a. carbon dioxide):
  `\text{CO}_2`.
• Molar mass of
  `\text{CO}_2`:
  `\underbrace{12.01}_{\text{C}} + 2*\underbrace{16.00}_{\text{O}}=44.01\;\text{g}\cdot\text{mol}^(-1)`.

• Formula of ethyne: structural
  `\text{H}-\text{C}\equiv\text{C}-\text{H}` or molecular
  `\text{C}_2\text{H}_2`.
• Molar mass of
  `\text{C}_2\text{H}_2`:
  `2*\underbrace{12.01}_{\text{C}}+2 *\underbrace{16.00}_{\text{O}} = 56.02\;\text{g}\cdot\text{mol}^(-1)`.

All carbon atoms in that 104 grams of ethyne will end up in
`\text{CO}_2`. Number of moles of molecules in 104 grams of ethyne:

`n = (m)/(M) = (104)/(56.02) = 1.85648\;\text{mol}`.

There are two carbon atoms in each ethyne molecule. Number of carbon atoms in that many ethyne molecules:

`n(\text{C}) = 2\;n(\text{C}_2\text{H}_2) = 3.71296\;\text{mol}`.

There are one carbon atom in each
`\text{CO}_2` molecule. In case oxygen is in excess, all those carbon atoms from that 104 grams of ethyne will make
`n(\text{CO}_2) = n(\text{C}) =3.71296\;\text{mol}` of
`\text{CO}_2`.

Mass of all those
`\text{CO}_2` molecules:

`m = n\cdot M = 163\;\text{g}`. (3 sig. fig. as in the mass of ethyne.)