Small quantities of oxygen can be prepared in the laboratory by heating potassiu

Question

Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KCIO_3(s). The equation for the reaction is 2KCIO_3 rightarrow 2KCl+3O_2 Calculate how many grams of O_2(g) can be produced from heating 58.7 grams of KCIO_3(s). Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric add with manganese(IV) oxide: 4HCl(aq) + MnO_2(s) rightarrow MnCl_2(aq) + 2H_2O(l) + Cl_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 CI_2? If the yield of the reaction is 85.1%, what is the actual yield of chlorine? 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

Part A : moles of KClO3 taken = 58.7 g/122.55 g/mol = 0.48 mols

mass of O2 produced = 0.48 x 3 x 32/2 = 23.04 g

Part B : moles of MnO2 = 43.7 g/86.94 g/mol = 0.50 mols

moles of HCl = 50.5 g/36.5 g/mol = 1.38 mols

(a) If all of MnO2 is consumed we would need = 0.50 x 4 = 2.0 mols of HCl

If all of HCl is consumed we would need = 1.38/4 = 0.345 mols of MnO2

Since moles of HCl is less than required,

Limiting reagent = HCl

(b) theoretical yield of Cl2 = 1.38 x 70.906/4 = 24.46 g

(c) If reaction yield is 85.1 %

Yield of Cl2 = 0.851 x 24.46 = 20.81 g

Part C : moles of propane = 15.2 g/44.1 g/mol = 0.345 mols

mass of O2 required for complete combustion = 5 x 0.345 x 32 = 55.20 g

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