Question

Question 33 of 60 Submit How many grams of precipitate will be formed when 20.5 mL of 0.800 M CO(NO3)2 reacts with 35.0 mL of 0.800 M NaOH in the following chemical reaction? CO(NO3)2 (aq) + 2 NaOH (aq) → Co(OH)2 (s) + 2 NaNO3 (aq)

Answer 1

20.5 mL of 0.8oom COCNO3)2 = 20.5 x 0.880 miimole = 16.4 milimole - 16.4 x 103 more. I 0.016 4 mole 0.800m Naoth - 35 x 0.800 minmole 35.0 m2 of = 28 miumole. 2 28.8 10 3 mole = 0.028 mole. co(NO3)2 (aq) + 2NaOH(aq) coloH) (s) 1 + 2 NaNo (9) So 1 mole co(NO3)₂ react with a mole Naot to precipitate 1 mole cocoH)2 . s. 0164 mole fully reacts when (2x 0.0164) mole. 2 0.0328 mole Naoth will present. But NaoH. is present 0.028 mole i've less than the required. So, Naot here plays the role of limiting mact i reagent and we have to calculate with respect on to Naohen 0.014 mole Naot reacts with dio7 mole co(NO₂) 0.028 mole Naoh reacts with 0.014 mole conoz) CO(NO3)2 + swoon Coco + 2 NaNoz moles a l a 0.014 - 0.014 0.028 0.014 moles color will precipitate molar mass of coloH)2 = 92.948 8/mole. 1 mole co(OH)2 = 92.948 ģ color% 0.014 mole coCOM)2 = 0.014 % 92.948 g CO (OH)2 = 1.3012 g co COH)2 .. 11.3012 g cocon will be precipitate ours