Question

Calculation of percent yield of [Co(NH₃)₅Cl]Cl₂:

I'm having trouble with these steps:

• Calculation of limiting reagent for synthesis of [Co(NH₃)₅Cl]Cl₂ (assume H₂O₂ present in excess)
• Calculation of theoretical yield of [Co(NH₃)₅Cl]Cl₂

These are the reactants we used: 2.010g NH₄Cl, 15mL 15M NH₃, 4.067g CoCl₂6H₂O, 3.2mL 32% H₂O_2, 15mL 12M HCl

This is the equation used: 2CoCl2*6H2O + H2O2 + 8NH3 + 2NH4Cl --> 2[Co(NH3)5Cl]Cl2 + 14H2O

Answer 1

2 loc.6 Hotho + 8 NH₃ + 2NH₂4 0.225 0. 225 0.037 0.03 tlumo 0.02 2 [LO (NY) U] 4₂ (LR) NHAU = 2.019 = 2 2 20.03757 mole. MN3 = 15x15x10 4.067 0.02 cool, -640 = 4.067g = 203 = 0.225 mote o-03 mole of m2 = 3.2 x 0-32 -= - 34 2 = 3.2x0.32g. HU = 15x12x103 R = 0.180 mole & here, coel-640 is limiting reaget, so yheld of [co (M43) 47ch, will be decided by Limiting reegut. so according to stoctione try of Reaction, mole of [Comu) ulk is 0.02 mole conch - 8.02 x 100 = o-bry.

Simply using mole concept we can get answer..