Question

With an H2SO4 concentration of 85% how many moles of water will be available to the reaction when 9.0 mL of 85% H2SO4 is used? (Density of 85% H2SO4 is 1.7 g/mL) calculations please This is used with the hydration of 1-hexene that has a density of 0.67 g/mL Also what is the theoretical yield in moles and grams of hexanol. Calculations please.

Answer 1

Density of 85% H2SO4 is 1.7g/mL. Thus, weight of 9.0 mL of 85% H2SO4 can be calculated as follows:

Wt. of 9.0 mL H2SO4 = Volume x density

Wt. of 9.0 mL H2SO4 = (9.0 mL) x (1.7 g/mL)

Wt. of 9.0 mL H2SO4 = 15.3 g

Thus, 15.3 g of 85% H2SO4 contains 15% water. Weight of water can be calculated as follows:

Wt. of water = Wt. of H2SO4 x (15/100)

Wt. of water = (15.3 g) x (15/100)

Wt. of water = 2.30g

Molar mass of water is 18.0 g/mol. Thus, number of moles of water in 2.30g of water can be calculated as follows:

Number of moles of water = (2.30 g)/ (18.0 g/mol)

Number of moles of water = 0.128 mol

Therefore, the number of moles of water available for the reaction is 0.128 mol

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The balanced equation of reaction of hydration of 1-hexene(C6H12) to hexanol (C6H14O) can be shown as follows:

C6H12 + H2O -> C6H14O

Thus, one mole of 1-hexene reacts with one mole of water molecule to give one molecule of hexanol. Therefore, 0.128 moles of water will give 0.128 moles of hexanol. Hence, theoretical yield of hexanol is 0.128 moles. Weight of 0.128 moles of hexanol can be calculated as follows: (Molar mass of 1-hexene is 84 g/mol and that of hexanol is 102 g/mol.)

Wt. of 0.128 mol of hexanol = moles x molar mass

Wt. of 0.128 mol of hexanol = (0.128 mol) x (102g/mol)

Wt. of 0.128 mol of hexanol = 13.1 g

Therefore, theoretical yield of hexanol is 13.1 g or 0.128 mol