1) Describe how to make 100.0 mL of a 0.500 mol/L MgCl2 solution.

Question

2) 404.2 g of ammonium nitrate is produced when 112.5 g of ammonia and excess nitric acid get combined. What is the percentage yield of the product?

Thanks for any and all of your help :)

Answer 1

1) Describe how to make 100.0 mL of a 0.500 mol/L MgCl2 solution.
How many moles do you want? That's mols = M x L = 0.500 M x 0.100 L = 0.0500 moles.
Then grams = mols x molar mass or g = 0.500 moles x molar mass MgCl2.
Transfer that many grams MgCl2 quantitatively into a 100 mL volumetric flask, add some distilled water, swirl to dissolve all of the MgCl2, add distilled water to the mark on the flask, stopper, mix thoroughly, label, done.

2) 404.2 g of ammonium nitrate is produced when 112.5 g of ammonia and excess nitric acid get combined. What is the percentage yielNH3d of the product?

NH3 + HNO3 ==> NH4NO3
mols NH3 = 112.5/molar mass NH3 = about 7 but you need a better answer than that.
1 mol NH3 will produce 1 mol NH4NO3 (look at the coefficients in the balanced equation). So about 7 mols NH3 will produce about 7 mols NH4NO3. Then grams NH4NO3 = moles NH4NO3 x molar mass NH4NH3. This is the theorectical yield (TY). The 404.2 g is the actual yield (AY).
% yield = (AY/TY)*100, Don't use the 7 from above but use the actual number you get and follow through with the rest of the problem using your numbers and not my numbers. Post your work if you get stuck.