Question

Consider the following reaction to answer the questions below. 2COCI-6H, 0(s) + 2NH, Cl(aq) +8NH, (aq) + H,0, (aq) -- 2[Co(NH2)HOCI,(s) + 12H 0(1) What is the percent yield if 3.25 g of (Co(NH3),H,O]CI, are produced when 4.4 g of Cocl, 64,0 reacts with 2.6 g of NH CI, 3.9 mL of 15 M NH,OHand 3.4 ml of 10.% H,0, (density 1.0 g/mL). Assume the reaction goes to completion (Do not include units in your answer. If you round during your calculations make sure to keep at least 3 decimal places. Report your answer to 1 decimal place.) Hint: 1 mole NH,(aq) - 1 mole NH,OH(aq) Answer:

Answer 1

Step 1: Find the limiting reagent.

a. Determine the number of moles of each reactant (n).

Molar mass of CoCl2.6H2O = 237.95 g/mol

Molar mass of NH4Cl = 53.5 g/mol

n = Given mass / Molar mass

n_CoCl2.6H2O = 4.4 g / 237.95 g/mol = 0.018 mol

n_NH4Cl = 2.6 g / 53.5 g/mol = 0.049 mol

n_NH4OH = Molarity x Volume in (L) = 15 M x 0.0039 L = 0.0585 g

n_H2O2 = 3.4 g / 34 g/mol = 0.1 mol

The one with the lowest value is the limiting reagent.

Hence, CoCl₂.6H₂O is the limiting reagent.

From the reaction we see that 1 mol CoCl₂.6H₂O produces 1 mol [Co(NH₃)₅H₂O]Cl₃

237.95 g CoCl₂.6H₂O produces 268.4 g [Co(NH₃)₅H₂O]Cl₃

Hence 4.4 g CoCl₂.6H₂O produces [(268.4 g / 237.95 g) * 4.4 g] = 4.96 g of [Co(NH₃)₅H₂O]Cl₃

Hence theoretical yield of [Co(NH₃)₅H₂O]Cl₃ = 4.96 g

Experimental yield = 3.25 g

Percent yield = (Experimental yield / Theoretical yield) * 100

= (3.25 g / 4.96 g) * 100

= 65.5%