Question

If 14.0 g of aluminum reacts with 39.4 g of oxygen gas how many grams of aluminum oxide will form?

If there is a 59% yield how many grams of aluminum oxide actually form?

Answer 1

Ans a) Let us first balance the chemical reaction of aluminium and oxygen gas

4Al + 3O₂ -> 2Al₂O₃

Now , Molar mass of Al = 27 g

Given mass of Aluminium = 14 g

Number of moles of Aluminium = Given mass /molar mass

                                              = 14 /27

                                              = 0.5185

Again molar mass of O₂ = 32 g

Given mass = 39.4 g

number of moles of O₂ = 39.4 / 32

                                  = 1.231

Here since we can see from the above reaction that 4 moles of Aluminium reacts with 3 moles of oxygen gas and from our calculation we can see that number of moles of Aluminium is less ,therefore here in the given question Aluminium is the limiting reagent and oxygen is present in excess.

Therefore in order to calculate the theoritical yield of aluminium oxide we can proceed as follows;-

= 0.2592 moles of Al₂O₃

Now , we will calculate the theoritical mass of aluminium oxide

Molar mass of Al₂O₃ = 101.96g

Number of moles of Al₂O₃ = 0.2592

Calculated mass = Number of moles x molar mass

                          = 0.2592 x 101.96

                          = 26.42 g

Therefore when 14 g of Al reacts with 39.4 g of O₂ it forms 26.42 g of Al₂O₃

b) Now ,

Given % yield = 59 %

Theoritical yield = 26.42 g

we know

% yield = (Actual yield /Theoritical yield) x 100

       Actual yield = 59x26.42 /100

                          =15.59g

Therefore for 59% yield , 15.59 g of Al₂O₃ actually formed.