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Ulleksa [173]
3 years ago
6

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

Chemistry
1 answer:
mars1129 [50]3 years ago
7 0

163 grams (3 sig. fig.).

<h3>Explanation</h3>
  • Formula of <em>carbon(IV) oxide</em> (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.)

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The complete ionic equation for the above  reaction would be;

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The spectator ions are;  Na⁺ and Cl⁻

The net ionic equation is given as;

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