Ethene gas (C2Ha) reacts with oxygen to form carbon monoxide and water. Write a balanced equation for this reaction, then find the mole ratios of substances on each side of the equation.
To write a balanced equation for the reaction between ethene gas (C2H4) and oxygen (O2), we need to determine the number of carbon, hydrogen, and oxygen atoms on each side of the equation.
First, we write the unbalanced equation:
C2H4 + O2 -> CO + H2O
Now, let's balance the equation step by step:
Balance the carbon atoms:
On the left side, we have 2 carbon atoms in C2H4, and on the right side, we have only 1 carbon atom in CO. To balance this, we put a coefficient of 2 in front of CO:
C2H4 + O2 -> 2CO + H2O
Balance the hydrogen atoms:
On the left side, we have 4 hydrogen atoms in C2H4, and on the right side, we have 2 hydrogen atoms in H2O. To balance this, we put a coefficient of 2 in front of H2O:
C2H4 + O2 -> 2CO + 2H2O
Balance the oxygen atoms:
On the left side, we have 2 oxygen atoms in O2, and on the right side, we have 4 oxygen atoms in 2CO and 2H2O. The total oxygen atoms on the right side are already balanced, so we don't need to make any changes.
Final balanced equation:
C2H4 + O2 -> 2CO + 2H2O
Now, let's determine the mole ratios of substances on each side of the equation:
The coefficients in a balanced equation represent the mole ratios of the substances. In this case, the mole ratio between C2H4 and O2 is 1:1, meaning for every 1 mole of C2H4, we need 1 mole of O2.
Similarly, the mole ratio between C2H4 and CO is also 1:1, and the mole ratio between C2H4 and H2O is 1:1.
Therefore, the mole ratios for this reaction are:
C2H4 : O2 : CO : H2O
1 : 1 : 2 : 2