please I urgently need the solution to this problem h. 5.0 moles of Fe and 4.0 m

Question

please I urgently need the solution to this problem

h. 5.0 moles of Fe and 4.0 moles of O e. 160 moles of Fe and 20.0 moles of O, For each of the following reactions, 20.0 g of each present initially, Determine the limiting reactant,d the grams of product in parentheses that would be a. 2Al(s) + 3Cl2(g)- 2AICI(s) (AICI.) b. 4NH,(g) + 50(x)- 4N0(g) + 6H,0(g) c. CS2(g) + 30,(8)-CO2(g) + 2SO2(g) (Soj_ 78 For cach of the following reactions, 20.0 g of each actan present initially. Determine the limiting reactant, and cal the grams of product in parentheses that would be prodi a. 4AI(s) + 302(g)- 2A1,03(s) (A120) b. 3NOj(g) + H2O(/)- 2HNo,(aq) + NO(g) ( c. GHOH(1) + 30,(g)--2002(g) + 3H0(1) For cach of the following reactions, calculate the g

Explanation / Answer

moles =mass/ molar mass , in case of Aluminium, moles =mass/atomic weight

Atomic weight: Al= 27, molar mass of Chlorine= 71

moles : Al= 20/27= 0.74, moles of Chlorine= 20/71= 0.28

the reaction is 2Al+3Cl2 --->2AlCl3(s)

Theoretical molar ratio of Al :Cl2= 2:3 = 1:1.5 actual molar ratio of Al :Cl2 =0.74:0.28= 1:0.28/0.74= 1:0.38

so Limiting is Cl2 and excess is Al. From the stoichiometry of the reaction, 3 mole of Cl2 gives 2 moles of AlCl3

0.28 moles of Cl2 gives 0.28*2/3= 0.1867 moles of AlCl3 while consuming (2/3)*0.28= 0.1867 moles of Al.

molar mass of AlCl3 =133.34 g/mole, mass of AlCl3= moles* molar mass = 0.1867*133.34 gm =24.89 gm

2,   moles of NH3= mass/molar mas=20/17= 1.176, moles of O2= 20/32= 0.625

the reaction is 4NH3(g)+ 5O2(g) ------>4NO(g)+6H2O(g), theoretical molar ratio of NH3:O2= 4:5= 1:1.25

actual molar ratio = 1.176:0.625 = 1:0.625/1.176= 1: 0.53

so limiting reactant is O2, since 1 mole of NH3 requires 1.25 noles of O2 where as supplied is only 0.53

5 moles of O2 gives 6 moles of water

0.625 moles of O2 gives 6*0.625/5= 0.75 moles of water. Molar mass of water= 18 g/mole, mass of water= moles* molar mass= 0.75*18=13.5 gm

3. molar masses: CS2= 76, moles of CS2= 20/76 =0.263, moles of O2= 20/32= 0.625

the reaction is CS2(g)+3O2(g) ----->CO2(g)+2SO2(g)

Theoretical molar ratio of CS2: O2= 1:3, actual molar ratio =0.263:0.625 = 1:0.625/0.263= 1:2.4

so limiting reactat is O2. 3 moles of Oxygen gives 2 moles of SO2

0.625 moles of Oxygen gives0.625*2/3= 0.42 moles of SO2, mass of SO2= moles* molar mass= 0.42*64= 26.88 gm

4. moles of Al = 20/27= 0.74, moles of O2= 20/32=0.625, the reaction is 4Al(s)+3O2(g)----->2Al2O3(s)

theoretical molar ratio of Al :O2= 4:3= 4/3 :1 = 1.33 :1 , actual molar ratio =0.74:0.625 =1.2:1

so limiting is Al, 4 moles of Al produces 2 mole of Al2O3

0.74 moles of Al produces 0.74*2/4 = 0.37 moles of Al2O3, molar mass of Al2O3= 102, mass of Al2O3= 102*0.37= 37.74 gm of Al2O3.

5. moles of NO2= mass/ molar mass= 20/46= 0.434, moles of H2O= 20/18= 1.111

the reaction is 3NO2(g)+ H2O(l) ------->2HNO3(aq)+ NO(g), molar ratio of NO2: H2O= 3:1 ( theoretical)

actual molar ratio = 0.434: 1.111 =0.434/1.111 : 1 = 0.39:1

so limiting reactant is NO2. 3 mole of NO2 gives 2 mole of HNO3

0.434 moles gives 0.434*2/3= 0.29 moles, molar mass of HNO3= 63, mass of HNO3= 0.29*63= 18.27 gm

6. moles of C2H5OH= 20/46= 0.43, moles of O2= 20/32= 0.625

the reaction is C2H5OH+ 3O2----->2CO2+ 3H2O

molar ratiof C2H5OH: O2= 1:3 ( Theoretical)

Actual molar ratio = 0.43:0.625= 1: 0.625/0.43= 1 :1.45

Oxygen is limiting. 3 moles of O2 gives 2 mole of CO2

0.625 moles of O2 gives 0.625*2/3= 0.42 moles of CO2 , mass of CO2 produced. =0.42*44= 18.48 gm

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