Question

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

Answer 1

163 grams (3 sig. fig.).

Explanation

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

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

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:

$n = \dfrac{m}{M} = \dfrac{104}{56.02} = 1.85648\;\text{mol}$.

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.)