Bismuth oxide reacts with carbon to form bismuth metal: Bi2O3(s) + 3C(s) → 2Bi(s) + 3CO(g)

When 447 g of Bi2O3 reacts with excess carbon,

(a) how many moles of Bi form?

(b) how many grams of CO form?

(a) 1.92 moles of Bi produced.

(b) 80.6 grams

Explanation:

             Balanced equation: Bi2O3(s) + 3C(s) → 2Bi(s) + 3CO(g)

1st find moles of Bi2O3:

Bi2O3 has Mr of 466 and mass of 447 g

2nd find moles of Bi:

Bi2O3 : 2Bi

→       1 : 2 ------ this is molar ratio.

→ 0.959227 : (0.959227)*2

→ 0.959227  : 1.91845

→ 0.959227  : 1.92

Therefore 1.92 moles of Bi was produced.

3rd Find moles of 3CO:

Bi2O3 : 3CO

      1 : 3

0.959227 : (0.959227 )*3

0.959227 : 2.87768

3CO has 2.87768 moles and we know the Mr is 28.

g

Therefore 80.575 grams of CO was produced.