When a mixture of 10 moles of SO2 and 15 moles of O2 was passed over a catalyst, 10 moles of SO3 was formed. how many moles of SO2 and O2 did not enter into combination?
So2=2 .O2=11
just use formula idiot
To determine the moles of SO2 and O2 that did not enter into the combination, we need to find the limiting reactant in the given chemical equation. The limiting reactant is the one that is completely consumed and determines the amount of product formed.
First, we need to identify the balanced chemical equation for the reaction:
SO2 + O2 --> SO3
From the equation, we can see that the stoichiometric ratio between SO2 and SO3 is 1:1, meaning that one mole of SO2 combines with one mole of O2 to form one mole of SO3.
Since 10 moles of SO3 were formed, it means that 10 moles of SO2 were consumed, assuming SO2 is the limiting reactant.
Now, let's find the remaining moles of SO2 and O2:
Initially, we had 10 moles of SO2. If we consumed 10 moles of SO2, we are left with 0 moles of SO2.
For O2, we started with 15 moles. Since the stoichiometric ratio between SO2 and O2 is 1:1, it means that we would need 10 moles of O2 to react with the consumed SO2.
Therefore, 10 moles of O2 reacted, and the remaining moles of O2 would be:
15 moles (initial) - 10 moles (reacted) = 5 moles of O2
In summary, 0 moles of SO2 and 5 moles of O2 did not react and did not enter into the combination.
2SO2 + O2 ==> 2SO3
The long way of solving but easiest to explain is this.
Convert 10 mols SO2 to mols SO3.
10 x (2 mols SO3/2 mols SO2) = 10
Convert 15 mols O2 to mols SO3.
15 x (2 mols SO3/1 mol O2) = 30 mols SO3.
So we know 10 mols SO3 is the limiting reagent. How much O2 will that use?
10 mols SO3 x (1 mol O2/2 moles SO3) = 5 mols O2. So we use all of the SO3(leaving zero not reacted), we use 5 of the 15 (leaving 10) moles O2 not reacted.