Combining 0.375 mol of Fe20.3 with excess carbon produced 10.6 g of Fe. Fe_2O_3
ID: 952958 • Letter: C
Question
Combining 0.375 mol of Fe20.3 with excess carbon produced 10.6 g of Fe. Fe_2O_3 + 3C rightarrow 2Fe + 3CO What is the actual yield of iron in moles? What was the theoretical yield of iron in moles? What was the percent yield? Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric add with manganese(IV) oxide: 4HCl(aq) + MnO_2(s) rightarrow MnCI_2(aq) + 2H_2O(l) + CI_2(g) You add 43.7 g of MnO_2 to a solution containing 50.5 g of HCI. What is the limiting reactant? What is the theoretical yield of Cl_2? If the yield of there reaction is 85.1%, what is the actual yield of chlorine? Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KCIO_3(s). The equation for the reaction is 2KClO_3 rightarrow 2KCl + 3O_2 Calculate how many grams of O_2(g) can be produced from heating 58.6 grams of KClO_3(s) The combustion of propane may be described by the chemical equation, C_3H_8(g) + 5O_2(g) rightarrow 3CO_2(g) + 4H_2O(g) How many grams of O_2(g) are needed to completely burn 15.2 g of C_3H_8(g)?Explanation / Answer
1.
The reaction is Fe2O3+3C --> 2Fe+3CO
1 mole of Fe2O3 gives 2 moles of Fe
0.375 moles of Fe2O3 gives 2*0.375= 0.75 moles of Fe
Since excess carbon is used, this is the theoretical yield.
Actual yield = Mass/Atomic weight =10.6/56=0.1893
% yield= 100* actual yield/theoretical yield= 100*0.1893/0.75 =25.24%
2,
The reaction is 2KClO3 ---> 2KCl +3O2
Molecular weights : KClO3=39+35.5+48= 122.5 KCl= 39+35.5= 74.5 and O2=32
So 2moles of KCl= 2*122.5= 245 gms and 3 moles of oxygen =3*32=96
2moles of KClO3 gives 3 moles of oxygen
245 gms of KClO3 gives 96 gms of oxygen
58.7 gms of KClO3 gives 58.7*96/245 gms of oxygen=23 gms of oxygen
3.
Molecular weight of MnO2= 87 and molecular weight of HCl =36.5 and Cl2= 71
The reaction is 4HCl+MnO2--->MnCl2+2H2O+Cl2
4 moles of HCl requires 1 mole of MnO2 to give 1mole of Cl2
Given mass of MnO2= 43.7gm and HCl = 50.5 gms
Moles of MnO2= 43.7/87=0.50 moles and Moles of HCl= 50.5/36.5=1.38
As per the stoichiometry, molar ratio of MnO2 : HCl = 4:1
As pe the given data, molar ratio of MnO2: HCl= 0.5: 1.38= 1:2.76
So HCl is the limiting reactant ( since requires is 4 and supplied is only 2.76)
4 moles of HCl gives 1mole of Cl2
1.38 moles of HCl gives 1.38/4= 0.345 moles of Cl2
Mass of Cl2= moles* Molecular weight = 0.345*71=24.495 gms
This is the theoretical yield.
Since the acutal yiled is 85.1%, mass of Cl2 =0.851*24.495=20.845 gms
4. Molecular weights : C3H8= 44 and oxygen= 32
moles of C3H8 in 15.2 gms =15.2/44=0.345 moles
As per the reaction , 1 mole of C3H8 requires 5 moles of oxygen
0.345 moles of C3H8 as per the stoichiometry requires 0.345*5 moles of oxygen=1.725 moles of oxygen
Mass of oxygen =moles* molecular weight = 1.725*32=55.2 gms