A mixture of 82.49 g of aluminum (26.98 g/mo and 117.65 g of oxygen (32.00 g/mol

Question

A mixture of 82.49 g of aluminum (26.98 g/mo and 117.65 g of oxygen (32.00 g/mol) is allowed react. Identify the limiting reactant and determine the mass of the excess reactant present in the vessel when the reaction is complete. 4Al_(s) + 3O_2_(g) gives 2Al_2O_3(s) O_2 is limiting: 19.81 g of Al remain O_2 is limiting: 35.16 g of Al remain Al is limiting: 16.70 g of O_2 remain Al is limiting: 35.16 g of O_2 remain Al is limiting: 44.24 g of O_2 remain What is the final concentration of HCI formed mixing 16.0 mL of 0.130 M HCI and 12.0 mL of 0.600 M HCI? 0.423 M 0.365 M 0.511 M 0.331 M none

Explanation / Answer

11) Answer) (e) Al is limiting; 44.24 g O2 remain.

Solution:

4 Al + 3 O2 -----------> 2 Al2O3

According to stoichiometry 4 mole of Al needs 3 mole of O2 for complete reaction.

4 mole of Al = 4×gram atomic mass = 4 mol × 26.98 g/mol = 107.92 g

3 mole of O2 =3 × 32.00 g/mol = 96.00 g

But in given problem ; mass of Al = 82.49 g & mass of O2 = 117.5 g

Limiting reagent is the reactant present in minimum quantity.The minimum quantity reactant determines amount of product formation and hence called limiting reagent.

So here limiting reagent is Al (s).

107.92 g Al needs 96 g O2for complete reaction.

Hence 82.49 g needs (96 ×82.49)/107.92 g O2 = 73.38 g

O2 is found excess in the reaction mixture and is = 117.5 - 73.38

= 44.2 g

So Answer) (e) Al is limiting; 44.24 g O2 remain.

12) Answer) ( D ) 0.331 M.

Molarity of of new solution obtained by mixing of two solutions of different molarity can be calculated using below formula.

( M1 × V1 ) + ( M2 × V2 ) = M3 × V3

Where ; M1= molarity of solution 1, V1= volume of solution 1

M2 =molarity of solution 2, V2 = volume of solution 2 )

M3 =molarity of new solution , V3 = volume of new solution =(V1+V2)

(16.0 ml×0.130 M)+(12.0 ml ×0.600 M) = M3 × (16.0+12.0)ml

M3 = 9.28/28 = 0.331 M

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