Question

.Rubidium nitrate reacts with cobalt (III) sulfide to yield rubidium sulfide and cobalt (III) nitrate

a. If two moles of cobalt (III) sulfide react with excess rubidium nitrate, how many grams of cobalt (III) nitrate will be produced?

b. How many grams of rubidium nitrate would be required to produce 6.83*1048 formula units of cobalt (III) nitrate?

c. If your experiment yielded 915.25 g of cobalt (III) nitrate, determine the percent yield?

Answer 1

The balanced reaction between rubidium nitrate (RbNO₃) reacts with cobalt (III) sulfide (Co₂S₃) can be written as follows.

6RbNO₃ + Co₂S₃ 3Rb₂S + 2Co(NO₃)₃

Part a.

From the balanced reaction, for 1 mole of Co₂S₃, 2 moles of Co(NO₃)₃ are produced.

Therefore, for 2 moles of Co₂S₃, 4 moles of Co(NO₃)₃ will be produced.

i.e. The mass of Co(NO₃)₃ = 4 mol * 244.945 g/mol = 979.78 g

Part b.

The no. of formula units of Co(NO₃)₃ produced = 6.83*10⁴⁸

Therefore, the no. of moles of Co(NO₃)₃ produced = 6.83*10⁴⁸/6.023*10²³ = 1.134*10²⁵ mol

i.e. The no. of moles of RbNO₃required = 3 * 1.134*10²⁵ mol * 147.472 g/mol = 5.017*10²⁷ g

Part c.

From part a, the theoretical yield of RbNO₃ = 979.78 g

Therefore, the percent yield = (915.25 g/979.78 g)*100 = 93.4%