b)(6.5 mol NO2) x (4/4) x (63.0130 g HNO3/mol) = 4.1 x 10^2 g HNO3-a) 2NH3 + H2SO4 -> (NH4)2SO4 .........
2. a) (Your empirical formula is correct.)
b) C2H4O = 44.0528 g/mol, so 88 would be twice that:
C4H8O2
3.b)
(6.5 mol NO2) x (4/4) x (63.0130 g HNO3/mol) = 4.1 x 10^2 g HNO3
a) 2NH3 + H2SO4 -> (NH4)2SO4 ... right
To figure out how many grams of aluminum sulfate, convert the given grams to moles
22.7 grams NH3 x 1mol NH3/17.03052 grams = 1.333 moles nh3
(grams divided by molar mass equals moles) There are 2 moles NH3 for every mole of aluminum sulfate so multiply 1.332901168 moles by (1/2) and you get .666 moles of (NH4)2SO4 multiply this by the molar mass of (NH4)2SO4 (132.13952) and get 88.06 grams
do the same for H2SO4
54.8 x 1/98.0784 x (1/1) x 132.13952 = 73.831 grams *****
The lesser value 73.8 grams is the answer, and Sulfuric acid was the limiting reactant because, had you not ran out of it, the reaction would have continued until you ran out of ammonia.
2. If you have found the answer to be C2H4O then find the molar mass by adding the atoms that make the molecule 2 carbons = 24.0214 , 4 hydrogens= 4.0316, 1 oxygen = 15.9994, altogether = 44.0524 ((((( now divide the given molar mass (88) by the one you found (44) to get 2. Multiply everything by this number (2) to get C4H8O2.
3. with all of the information that I have given you, you should be able to figure this one out on your own (you have to work it backwards) if given moles use the balanced eqn to figure out the moles of the other side of the eqn. To go from moles to grams, multiply by the molar mass, To go from grams to moles (you wont have to in this problem) divide by the molar mass
