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.

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

passing a magnet thought the

mixture , because iron gets attracted to magnet .