Answer:
a) 1.44 kg
b) 1.47 kg
Explanation:
a) By the reaction given, the stoichiometry, the molar ratio, is:
2 moles of C3H6 : 2 moles of NH3 : 3 moles of O2 : 2 moles of C3H3N : 6 moles of H2O, or 2:2:3:2:6.
By the mixture given, one or two of the reactants may be in excess, so, we must found which of them is the limiting reactant, the reactant that will be totally consumed.
The molar masses of the compounds are:
C3H6 = 42.0 g/mol
NH3 = 17.0 g/mol
O2 = 32.0 g/mol
C3H3N = 53.1 g/mol
H2O = 18.0 g/mol
Thus, the mass ratio between the reactants will be the molar ratio multiplied by the molar mass:
2 moles of C3H6*42.0 g/mol --- 2 moles of NH3*17.0 g/mol -- 3 moles of O2* 32.0 g/mol
84.0 g of C3H6 -- 34.0 g of NH3 -- 96.0 g of O2
Thus, assuming C3H6 as limiting, let's use the rule of three to find out which would be the masses of the other reactants needed:
84.0 g of C3H6 -- 34.0 g of NH3
1.14 kg -- x
84.0x = 38.76 kg
x = 0.46 kg of NH3
Because there's more NH3 than it's required, NH3 is in excess, and C3H6 is limiting. Let's test using O2 as reference:
84.0 g of C3H6 -- 96.0 f of O2
1.14 kg -- x
84.0x = 109.44
x = 1.30 kg of O2
So, O2 is also in excess, and the limiting reactant is C3H6.
Thus, the molar ratio between C3H6 and C3H3N is 2:2, which is simplified by 1:1, and so the mass ratio is 42 g of C3H6 -- 53.1 g of C3H3N, so the mass of acrylonitrile can be found by a rule of three:
42 g of C3H6 -- 53.1 f of C3H3N
1.14 kg -- x
42x = 60.534
x = 1.44 kg
b) The molar ratio between C3H6 and water is 2:6, which can be simplified by 1:3, thus the mass ratio is 1*42g/mol of C3H6 -- 3*18 g/mol of H2O
42 g of C3H6 -- 54 g of H2O
By a rule of three the mass of water is:
42 g of C3H6 -- 54 g of H2O
1.14 kg -- x
42x = 61.56
x = 1.47 kg
