What volume of 1.20 M nitric acid is needed to convert 5.5g of m-chlorophenol to 3-chloro-2,6-dinitrophenol?
I So far I have converted the 5.5g chlorophenol to mols and I got .043 mol. I know Volume * Molarity (1.2M)=moles. At this point I am stuck.
Writing the real subject title will usually get you answers sooner.
Using the coefficients in the balanced equation, convert mols chlorophenol to mols of HNO3 need to make the product. Then you know M = moles/L; L is the only unknown.
DrWLS is right. I wouldn't click on ORGO but if you wrote chemistry there for the subject you would have obtained a quicker answer.
Thank you I was able to work it out:
it's a 1:1 ratio so since 5.5g chlorophenol is .043 mols, nitric acid must be .043 mols as well.
V(1.20M) = .043mol
V= .036L
To determine the volume of 1.20 M nitric acid needed to convert 5.5g of m-chlorophenol to 3-chloro-2,6-dinitrophenol, you are on the right track. Let's continue with the calculation.
1. Start by converting the mass of m-chlorophenol to moles. You have correctly calculated it as 0.043 mol.
2. The balanced chemical equation for the reaction between m-chlorophenol and nitric acid can give us the stoichiometric ratio. Without the equation, it is not possible to calculate the volume of nitric acid required accurately. Do you happen to have the balanced chemical equation for the reaction?
Once you provide the balanced equation, we can proceed with the calculation by using the stoichiometric ratio and the molarity of nitric acid.