24 Give the percent yield when 2751 g of Co2 with 4.000 moles of O2 e percent yi

Question

24 Give the percent yield when 2751 g of Co2 with 4.000 moles of O2 e percent yield when 27.51 g of Co2 are formed from the reaction of 4.000 moles of C8H18 2C8H18 + 2502 - 16 CO2 + 18H20 (A) 20.00% (B) 25.00% (C) 50 00% (D) 12.50% 25. Choose the reaction that represents the combustion of CgH1202 (A) C8H12020) + 8 O2g) 6 CO2g) + 6 H2O(g) (B) Mg(s) + C6H120210- MgC6H12O2(aq) (C) 6 C(s) + 6 H2(g) + O2(g) -- C6H120210 (D) C6H12O2(0) -6 C(s) + 6 H2(g) + O2(g) (D) Rb 26. Which should give the least vigorous reaction when dropped in water? (A) LI (B) Na (C) K 27. Why are the halogens among the most active nonmetals? (A) Their valence electrons are more effectively shielded. (B) All halogens have low electron affinities (C) Their d orbitals are completely filled. (D) They only need one electron needed to attain a nobel gas configuration 28. How many moles of nitrogen are formed when 58.6 g of KNO3 decomposes according to the following reaction? (The molar mass of KNO3 is 101.11 g/mol). 4 KNO3(s) + 2 K20(s) + 2 N2(g) + 5 O2(g) (A) 0.290 mol N2 (B) 0.580 mol N2 (C) 18.5 mol N2 (D) 0.724 mol N2 29. Determine the limiting reactant (LR) of nitrogen gas that can be formed from 50 og N2O4 and 45.0g. N2H4. Useful molar masses are: N2O4 = 92.02 g/mol, N2H4 = 32.05 g/mol. N204n + 2 N2H40 -- 3 N2 + 4 H2O (A) LR = N2H4 (B) LR = N204 30. Determine the theoretical yield of HCI if 60.0g of BCl3 and 37.5 g of H20 are reacted according to the following balanced reaction. A possibly useful molar mass is BCl3 = 117.16 g/mol. BCl3(g) + 3 H2O(l) H3BO3(s) + 3 HCl(g) (A) 75.9 g HCI (B) 132 g HCI (C) 187 g HCI (D) 56.0 g HCI

Explanation / Answer

24)

we have the Balanced chemical equation as:

2 C8H18 + 25 O2 ---> 16 CO2 + 18 H2O

2 mol of C8H18 reacts with 25 mol of O2

for 4 mol of C8H18, 50 mol of O2 is required

But we have 4 mol of O2

so, O2 is limiting reagent

we will use O2 in further calculation

Molar mass of CO2 = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

From balanced chemical reaction, we see that

when 25 mol of O2 reacts, 16 mol of CO2 is formed

mol of CO2 formed = (16/25)* moles of O2

= (16/25)*4

= 2.56 mol

we have below equation to be used:

mass of CO2 = number of mol * molar mass

= 2.56*44.01

= 1.127*10^2 g

% yield = actual mass*100/theoretical mass

= 27.51*100/1.127*10^2

= 24.4 %

Answer: B

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