How many moles of dichloromethane (CH2Cl2) would be produced if we doubled the amount of chlorine, but still had the same amount of ethene (C2H4) available in the reaction?
You must describe what you are doing before we know how to answer. Probably this is a limiting reagent problem.
To determine the number of moles of dichloromethane (CH2Cl2) produced, we need to consider the balanced equation for the reaction between chlorine (Cl2) and ethene (C2H4) to form dichloromethane.
The balanced chemical equation for this reaction is:
C2H4 + Cl2 -> CH2Cl2
From the balanced equation, we can see that one mole of ethene reacts with one mole of chlorine to produce one mole of dichloromethane. Therefore, the stoichiometric ratio between ethene and dichloromethane is 1:1.
Now, if we double the amount of chlorine, it means we would have twice the number of moles of chlorine available for the reaction.
So, if we initially had "x" moles of ethene and "y" moles of chlorine, the reaction would consume "x" moles of ethene and "x" moles of chlorine to produce "x" moles of dichloromethane.
If we double the amount of chlorine, we would have "2y" moles of chlorine available. However, since the stoichiometric ratio between ethene and dichloromethane is 1:1, the maximum moles of dichloromethane that can be produced will still be "x" moles.
Therefore, doubling the amount of chlorine does not change the amount of dichloromethane produced. The number of moles of dichloromethane would remain the same as before.