B N J H J J H 19. Copper nanoparticles are produced by the reaction: CuBr2 + 2Na
ID: 928841 • Letter: B
Question
B N J H J J H 19. Copper nanoparticles are produced by the reaction: CuBr2 + 2NaBH, 2NaBr + Cu + H2+ B2H6. How much copper is produced if 0.25 g of copper(I) bromide react with excess NaBH,? (A) 0.035g (C) 0.11 g (B) 0.071 g (D) 0.14 g 20. What is the percent yield if 5.00 g of potassium react with excess oxygen to produce 4.25 g of solid K20? (A) (C) 4K(s) +O2(g) 2K20(s) (B) 35.3% (D) 85.0% 17.6% 70.6% If 25.00 mL of aqueous barium hydroxide is neutralized with 15.35 mL of 0.1350 M HCI, what is the concentration of the Ba(OH)2 solution? 21. Ba(OH)2(aq) + 2HCl(aq) BaC12(aq) + 2H20(1) (A) 0.04144 M (C) 0.1099 M (B) 0.08289 M (D) 0.1658 MExplanation / Answer
19) As per the balanecd reaction, 1 mole of CuBr2 on complete reaction produces 1 mole of Cu
Thus, molar mass of CuBr2 = 223.3 g/mole
Thus, moles of CuBr2 in 0.25 g of it = mass/molar mass = 0.25/233.3 = 1.12*10-3
Thus,moles of Cu produced = 1.12*10-3
mass of Cu produced = moles*molar mass = 1.12*10-3*63.5 = 0.071 g
20) molar mass of K = 39 g/mole
Thus, moles of K in 5 g of it = mass/molar mass = 5/39 = 0.128
Now, as per the balanced reaction , 4 moles of K on complete reaction produces 2 moles of K2O theoretically
Thus, theoretical moles of K2O produced on reaction of 0.128 moles of K = 0.064
Theoretical mass of K2O produced = moles*molar mass = 0.064*94 = 6.026 g
Thus, % yield = (actual mass of K2O/theoretical mass of K2O)*100 = 70.53%
Hence the correct option is :- (c)
21) Moles of HCl used in neutralization = molarity*volume of HCl olution in litres = 0.135*0.01535 = 2.072*10-3
Thus, as per the balanced reaction , moles of Ba(OH)2 neutralized = (1/2)*moles of HCl = 1.036*10-3
Thus, molarity of Ba(OH)2 solution = moles of Ba(OH)2/Volume of solution in litres = (1.036*10-3)/0.025 = 0.04144 M
Hence, the correct option is:- (A)
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