How to Calculate Mass of C6H5NO2 from Given Reactants?

[Aya]

the folowing reaction proceeds with a 70% yeild

C6H6 + HNO3 = C6H5NO2 + H2O

calculate the mass of C6H5NO2 expected if 12.8g of C6H5 reacts with excess HNO3

How do you do this problem?

 

[finchie_88]

I'm no expert, but this was what I think:

The number of moles of C6H6 can be calculated (mass/Mr), and since 1 mole of reactant produces 1 mole of product and the reaction is only 70% "efficient", means that the number of moles of the product is 0.7 times the number of moles of the C6H6, and so from this, simply multiply by the Mr of the C6H5NO2 to get the result.

btw, the excess bit tells you that all 12.8g of C6H6 reacts.

 

[Aya]

^ Thanks

#mols of C6H6

n=m/M
n=12.8g/78.06g/mol
n=0.164mol


#mols of C6H5NO2

=0.164mol(0.70)
=0.1148mol

mass of C6H5NO2

m=nM
m=0.1148mol(123.06g/mol)
m=14.1g

Like this?

 

[Aya]

and I have another question

If 3.45 x 10^23 formula units of Hg(NO3)2 are reacted with excess of Na2S, what mass of HgS can be expected if this process occurs with 97% yield?What are fomula units?

you would do that question the same way as the previous one right?