Question

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?

Answer 1

1.92 moles of Bi are formed

80.6 grams CO are formed

Step-by-step explanation:

You are given 447 g Bi2O3 (from the periodic table add up the atomic mass of 2 Bi + 3 O to get 465.96 grams Bi2O3 = 1 mole) and the ratio from the balanced equation is 1 mole Bi2O3 = 2 mole Bi. now you have all ratios you can set up your problem:

447 g Bi2O3 x 1 mole Bi2O3 x 2 mole Bi = 1.92 mole Bi

465.96 g Bi2O3 1 mole Bi2O3

Looking at the balanced equation you find the ratio 3 mole CO = 1 mole Bi2O3, and from the periodic table we know C is 12 g/mole and O is 16 g/mole (12 + 16 = 28 grams / mole):

447 g Bi2O3 x 1 mole Bi2O3 x 3 mole CO x 28 g CO = 80.6 grams CO

465.96 g Bi2O3 1 mole Bi2O3 1 mol CO

Answer 2

(a) 1.92 moles of Bi produced.

(b) 80.6 grams

Step-by-step 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

`\hookrightarrow moles = (mass)/(Mr)`

`\hookrightarrow moles = (447)/(466)`

`\hookrightarrow moles = 0.959227`

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.

`\hookrightarrow mass = moles * Mr`

`\hookrightarrow mass = 2.87768 * 28`

`\hookrightarrow mass =80.575` g

Therefore 80.575 grams of CO was produced.