Question

1. In the reaction between glucose and oxygen below, how many grams of glucose is required to produce 12.2 L of carbon dioxide if the density of Carbon dioxide is 1.26 g/L, C2H5OH, is 0.789g/mL.

C6H12O6(s) + 6 O2(g) = 6 CO2(g) +6 H2O

2. When 10.6 g of boron trichloride is reacted with water, 5.12 g of Hydrochloric acid is produced. Using the following balanced equation, calculate the percent yield for HCl for the reaction? (BCl3 = 117.17 g/mol, H2O = 18.02 g/mol, HCl = 36.46 g/mol and B(OH)3 = 61.93 g/mol.

BCl3(g) + 3 H2O (l) = 3 HCl(aq) + B(OH)3(aq)

Answer 1

1)

C6H12O6(s) + 6 CO2(g) ---------------- 6 CO2(g) + 6 H2O(l)

volume of CO2 = 12.2L

density of CO2 = 1.26 g/L

i,e mass of 1L of CO2 = 1.26 grams

mass of 12.2 L of CO2 = ?

                                      = 1.26 x 12.2/1 = 15.372 grasm

mass of CO2 = 15.372 grams

molar mass of CO2 = 44.0 gram/mole

molar mass of C6H12O6 = 180.16gram/mol

according to balanced equation

6 moles of CO2 = 1 mole of C6H12O6

6 x 44.0 grams of CO2 = 180.16 grams of C6H12O6

15.372 grams of CO2 = ?

                                   = 180.16 x 15.372/6x44.0 = 10.49 grams

mass of glucose required = 10.49 grams.

2)

BCl3(g) + 3 H2O(l) --------------- 3 HCl(aq) + B(OH)3(aq)

mass of BCl3 = 10.6 grams

molar mass of BCl3 = 117.17g/mole

molar mass of HCl = 36.46 g/mole

according to equation

1 mole of BCl3 = 3 mole of HCl

117.17 grams of BCl3 = 3 x 36.46 grams of HCl

10.6 grams of BCl3 = ?

                              = 3 x 36.46 x 10.6/117.17 = 9.89 grams

mass of HCl formed = 9.89 grams

Theoritical yield of HCl = 9.89 grams

Actual yield of HCl = 5.12 grans

Percent yield = Actual yield/theoritical yield x100

Percent yield = 5.12/9.89 x100 = 51.77%

percent yield of HCl = 51.77%.