Explanation:

c = Concentration of the solution
n = Moles of compound in solution
V = Volume of the solution
a) 2.00 L of 18.5 M of concentrated sulfuric acid.
n= ? c = 18.5 M, V = 2.00 L

n = 37 moles of sulfuric acid
b) 100.0 mL of
of sodium cyanide
n= ? ,c =
, V = 100.0 mL = 0.1 L

n =
of sodium cyanide
c) 5.50 L of 13.3 M of concentrated formaldehyde.
n= ? c = 13.3 M, V = 5.50 L

n = 73.15 moles of formaldehyde.
d)325 mL of
of iron sulphate
n= ? ,c =
, V = 325 mL = 0.325 L

n =
of iron sulfate.