Question

2. For each of the following, use dimensional analysis to perform the requested calculation: a. How many grams of water will be in 35.85 g CuSO4 3 H2O?! b. A student needs 125.00 g Al(NO3)3. The only bottle available is a hydrate: AI(NO3)3 9 H20. How many grams of hydrate must the student weigh out to get 125.00 g Al(NO3)3?

Answer 1

Here is the Solution:

(a). We need to find out the mass of water in 35.85 g of CuSO4:3H20. - First find out the molar mass of CuSO4:3H20 and water (H20) (i). Molar msss of CuSO4:3H20 there are 1Cu, 15, 70, 6H in CuSO4:3H2O Cu = 63.546 g/mol S= 32.065 g/mol O= 15.999 g/mol H= 1.008 g/mol Now, [63.546 +32.065 +(15.999 x 7)+(1.008 6)]g/mol = (63.546 +32.065 + 111.993 +6.048) g/mol = 213.652 g/mol (ii). Molar mass of H2O there are 2H and 10 in H2O H= 1.008 g/mol O= 15.999 g/mol Now, [(1.008* 2) + 15.999] g/mol = (2.016 + 15.999) g/mol = 18.015 g/mol Now we have 3 mois of water (H2O) in the given formula. So, the mass of water in that formula = 3 x 18.015 g/mol = 54.045 g/mol 54.045 almol Calculate the percent of water by mass =- x 100% = 25.2958% 213.652 g/mol So, we have 3 g of water in 25.2958% 25.2958% x3 = 0.75887 gram water. Hence, the mass of water is: 0.76 gram. Therefore we have, 100% (b). We know that, the molar mass of water and Aluminium nitrate are: AI(NO3)3 = 212.996 g/mol H2O= 18.02 g/mol Now, we need to find out the how many grams of hydrates must the student weigh out to get 125.00 g AI(NO3)3. we only have hydrate, AI(NO3)3 - 9H20 So, the molar mass of hydrate will be: molr mass of Al(NO3)3 +9 x molar mass of H20) = [212.996 +(9 x 18.02)]g/mol = (212.996 + 162.18) g/mol = 375.17 g/mol So, the student must weigh out, 125.00 g Al(NO3)x 375.17 g hydrate Al(NO3)3.9H20/mol 212.996 g AI(NO3)3/mol 125.00 g AlfhezTz x 375.17 g hydrate Al(NO3)3.9H20/mol 212.996 g AlfH633/m01 = 220.17 g hydrate) AI(NO3)3-9H20

Thanks! :)