Question

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Example: HW When aqueous solutions of silver nitrate and potassium dichromate are mixed, an exchange reaction occurs, and silver dichromate is obtained as a red solid. The overall chemical equation for the reaction is as follows: 2AgNO3(aq)+K2Cr2O3(aq)-Ag,Cr,0,($)+2KNO3(aq) What mass of Ag2Cr207 is formed when 500 mL of 0.17 M K2Cr207 are mixed with 250 mL of 0.57 M AgNO3? Hint: • Calculate No. of moles of both reactants from given Molarity and Volumes • See what is limiting reagent from two reactants • Use the no. of moles of limiting reagent to calculate mass of K,Cr,O, formed for this reaction.

Answer 1

- Age61207 + 2KNO3 2 AgNO3 + K2 Cr 2 Of (aq) Ag2 Cr 207 formed. To calculate Mass of Given, 500ml of 250 ml of 0.17M 0.57 M Ke Cr2O7 AgNO3. Molecular K2Cr2O7 = 294• 185 g/mol AgNO₃ = 169.87 glmol. AgaCr 207 = 431.76 g/mol equivalent gives 3 2 AgNO3 = 1 K2Cr2O7 = 1 Ag 2 Cr 207 Molecelas Weight 2 x 169.87 1 x 294.185 = 1 X431.76 lets calculate solution. mass of K2Cr2O7 in 500ml of 0.17M. 294.1859 in 1000ml gives ag in sooml in 1M 0.17M x = 294.185 X 500 X 0.17. 1000 a= 25.0g

Lets calculate 0.57 M mass of AgNO3 in 250 ml of AGNO3 Solution It. 169.879 in ag in 1000ml 250ml gives gives YM. 0.57M = 169.87 x 250 X 0.57 1000 m = 24-2 g Lets gee ï (both weactants are és oliy equivalent 2 moles Of AgNO3 Deedo mole of K2 Cr 204 i 2x169.87 = 294.185 - 24.2 = ng = a g. 2 2 20.99 Therefore, amount of K2Cr2O7 required for a 24-2g of AgNO- ris 20.9 g and not 25.0 g K2C207 is taken in exces quantity there are mass of Agzch207 depends upon Agno mass

2 moles of AgNO3 gives I mole of Ag₂Cr₂O7 .: 2x 169.87 24.2 g = 43176 almol - ag. 24.2 X 431 76 169.87 x 2 = 30.75 g mass of Ag2rr207 obtained is 30.75 g.