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

Answer:

(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

$\hookrightarrow moles = \frac{mass}{Mr}$

$\hookrightarrow moles = \frac{447}{466}$

$\hookrightarrow moles = 0.959227$

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.

Answer 2

Answer:

1.92 moles of Bi are formed

80.6 grams CO are formed

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