A particular compound contains, by mass, 25.5% carbon, 40.4% fluorine, and 34.1%
ID: 1073696 • Letter: A
Question
A particular compound contains, by mass, 25.5% carbon, 40.4% fluorine, and 34.1% oxygen. A 0.050-mol sample of this compound weighs 5.80 g. The molecular formula of this compound is C_6F_6O_6+ C_4F_4O_4- CFO_- C_3F_3O- C_3F_3O_3- 2AI(s) + 6HCI(aq) rightarrow 2AICI_3(aq) + 3H_2(g) According to the equation above, how many grams of aluminum are needed to completely react with 5.14 mol of hydrochloric acid? 62.5 g 138.7 46.2 g 5.14 g 416 g When 20.0 g C_2H_6 and 60.0g O_2 react to form CO_2 and H_2O, how many grams of water are formed? 20.0 g 18.0 g 58.0 g 14.5 g none of these A 5.95-g sample of AgNO_3 is reacted with BaCl_2 according to the equation 2AgNO_3(aq) + BaCl_2(aq) rightarrow 2AgCI(s) + Ba(NO_3)_2(aq) to give 3.22 g of AgCI. What is the percent yield of AgCl? 42.8% 54.1% 32.1 100% 64.1%Explanation / Answer
element % A.Wt Relative number simple ratio
C 25.5 12 25.5/12 = 2.125 2.125/2.125 =1
F 40.4 19 40.4/19 = 2.126 2.126/2.125 =1
O 34.1 16 34.1/16 = 2.13 2.13/2.125 = 1
Empirical formula = CFO
molar mass = mass/no of moles
= 5.8/0.05 = 116
molecular formula = ( empirical formula)n
n = M.Wt/E.F.Wt
= 116/47 = 3
d. C3H3O3
16. 2Al + 6HCL -------> 2AlCl3 + 3h2
6 moles of HCl react with 2 moles of Al
5.14 moles of HCl react with = 2*5.14/6 = 1.713 moles of Al
mass of Al = no of moles * gram atomic mass
= 1.713*27 = 46.25G >>>>>ANSWER c
17. 2C2H6 + 5O2 --------> 4Co2 + 6H2O
no of moles of C2H6 = W/G.M.Wt = 20/30 = 0.66 moles
no of moles of O2 = 60/32 = 1.875 moles
2 moles of C2H6 react with 5 moles of O2
0.66 moles of C2h6 react with = 5*0.66/2 = 1.65 moles of O2
C2H6 is limiting reagent
2 moles of C2H6 react with o2 to gives 6 moles of H2O
0.66 moles of C2H6 react with O2 to gives = 6*0.66/2 = 1.98 moles of H2O
mass of H2O = 1.98*18 = 35.64g
e. none of these
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