Problem 67E

What is the minimum amount of 6.0 M H2SO4 necessary to produce 25.0 g of H2 (g) according to the reaction between aluminum and sulfuric acid?

Solution: Here, we are going to calculate the mass of ethanol required to make 1.00 L of vodka. Also, we are going to calculate the volume of ethanol required to make 1.00 L of vodka. Step1: Molarity is defined as the number of moles of the solute(Here ethanol is the solute) in 1 litre of the solution. Mathematically, Number of moles of solute Molarity(M) = -------------------------------------- ------(1) Volume of solution in litres Step2: a) Given, molarity of the ethanol solution = 6.86 M Volume of the ethanol solution in Litres = 1.00 L Using the mathematical expression for molarity calculation, we get, number of moles of ethanol = Volume of solution in L X Molarity ethanol = 1.00 l X 6.86 M = 6.86 mol. Now, 1 mole of ethanol(CH CH3H) 2molar mass of ethanol Therefore, 6.86 mol of ethanol = 6.86 X molar mass of ethanol = 6.86 X 46.0682 g [molar mass of CH 3 2 = 46.0682 g] = 316.028 g Thus, 316.028 grams of ethanol should be dissolved in water to make 1.00 L of vodka. Step3: b) Density can be defined as mass per unit volume. i.e., Density = mass / volume Therefore, volume = mass / density Here, mass of ethanol = 316.028 g Density of ethanol = 0.789 g/mL Therefore, volume of ethanol = 316.028 g / 0.789 (g/mL) = 400.54 mL Thus, 400.54 mL of ethanol is required to make 1.00 L of vodka. ------------------