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of 600 g of aluminum oxide with 30.0 g of carbon produced 225 g of aluminum. Wha

ID: 543653 • Letter: O

Question

of 600 g of aluminum oxide with 30.0 g of carbon produced 225 g of aluminum. What is the percent for th reaction 31.8% C)141 % D)50.1% E)70.9% 18. Identify the major ionic species present in an aqueous solution of CoH1206 (glucose). A) 6 C, 12 H,60 C6 CH2,6o2 E) no ions are present 19. Based on the solubility rules, which one of the following compounds should be insoluble in water? A) NaCI B) MgBr2 C) FeCl2 D) AgBr E) ZnCl2 20. Identify the precipitate(s) formed when solutions of Na2SO4(aq, Ba(NO3)2(aq), and NH4C4are m xed A) Ba(C1O4)2 B) BaS04 and (NH42804 C) NH4NO3 and NaCIO4 D) BaSO4 E) NaC104 21. Identify the correct net ionic equation for the reaction that occurs when solutions of A) AgNO3(aq) + NH4C1(aq) AgCl(s) + NH4Cl(aq) AgNO3 and NH4CI are mi C) AgNO3(aq) + NH4Cl(aq) Agci(s) + NH4Cl(s) D) Ag+(aq) + cr(aq) AgCl(s) E) AgNOs(ag)+NH4 ()- NH4AgNO3(8) 22. For which one of the following acids is chlorine in the +5 oxidation state? A) HCI B) HCIo C) HCI02 D) HCIO3 E) HCIO4

Explanation / Answer

17)

Molar mass of Al2O3,

MM = 2*MM(Al) + 3*MM(O)

= 2*26.98 + 3*16.0

= 101.96 g/mol

mass(Al2O3)= 60.0 g

number of mol of Al2O3,

n = mass of Al2O3/molar mass of Al2O3

=(60.0 g)/(101.96 g/mol)

= 0.5885 mol

Molar mass of C = 12.01 g/mol

mass(C)= 30.0 g

number of mol of C,

n = mass of C/molar mass of C

=(30.0 g)/(12.01 g/mol)

= 2.498 mol

Balanced chemical equation is:

Al2O3 + 3 C ---> 2 Al + 3 CO

1 mol of Al2O3 reacts with 3 mol of C

for 0.5885 mol of Al2O3, 1.7654 mol of C is required

But we have 2.4979 mol of C

so, Al2O3 is limiting reagent

we will use Al2O3 in further calculation

Molar mass of Al = 26.98 g/mol

According to balanced equation

mol of Al formed = (2/1)* moles of Al2O3

= (2/1)*0.5885

= 1.1769 mol

mass of Al = number of mol * molar mass

= 1.177*26.98

= 31.75 g

% yield = actual mass*100/theoretical mass

= 22.5*100/31.75

= 70.9 %

Answer: E

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