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Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide:
4HCL(aq) + MnO2(s) ----> MnCl2(aq) + 2H2O(l) + Cl2(g)
You add 42.5 g of MnO2 to a solution containing 47.7 g of HCl.
(a) What is the limiting reactant? MnO2 or HCL?
(b) What is the theortical yield of CO2?
(c) If the yield of the reaction is 79.9%, what is the actual yield of chlorine?

Respuesta :

busyfingers1193

Answer:

a) HCl

b) 22.9g

c) 18.11g

Explanation:

MMn = 54.94g/mol

MO2 = 2(16) = 32g/mol

MH = 1g/mol

MCl = 35.45g/mol

Molar Mass of MnO2:

54.94 + 2(16)= 86.94

Molar Mass of HCl:

1+35.45=36.45

Mols of MnO2:

42.7 /86.94 = 0.49

Mols of HCl:

47.1 /36.45 = 1.29

Molar Mass of Cl2:

35.45 ×2 = 70.9

Mols of Cl2

1.29/4=0.323

Mass of Cl2 (Theoretical yield)

0.323 ×70.9= 22.9

To calculate the actual yield, we multiply the theoretical yield by the final percentage:

22.9 ×0.791=18.11

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