What is the coefficient of H20 when the following equation is properly balanced

Question

What is the coefficient of H20 when the following equation is properly balanced A)3 B) 4 C) 6 D) 12 E) 24 14. Chloribe gas can be made from the reaction of manganese dioxide with hydrochloric acid )+4liciaq) MaCl2(aq) + 2112001) +Cl2(g) to the above reaction, determine the limiting reactant when 5.6 moles of Mn02 are reacted wisth 7.5 moles of A) Mn02 B) HCI C) MaClh D) Cl2 E) No reagent is limiting 15. Calorine gas reacts with phosphorus to produce phosphorus pentachloride. How many grams of PCIs are produced from 3.5 g of Cih and excess P? A) 1.4g B)4.1 g C) 8.2 g D) 0.020 g E) 730 g 16. Calculate the mass of excess reagent remaining at the end of the reaction in which 90.0 g of SOz are mixod with 100 g of 02 2S02 +02 2503 A) 115g B) 225 g )675g D) 775g E) 400 g

Explanation / Answer

13) the balanced chemical equation will be

Al4C3   + H2O ---> Al(OH)3 + CH4

Al4C3   + 12H2O ---> 4Al(OH)3 + 3CH4

so coeffecient of H2O is 12

14)

The balanced equation is

MnO2 + 4HCl ---> MnCl2 + 2H2O + Cl2

So here 1mole of MnO2 will react with four moles of HCl

Therefore for 5.6moles of MnO2 we need = 4 X 5.6 moles of HCl = 22.4 moles

The moles of HCl present = 7.5 moles

So the limiting reagent is HCl

15) The reaction is

5Cl2 + 2P ---> 2PCl5

As per the balanced equation 5 moles of Cl2 will give two moles of PCl5 when Phosphorous is taken in excess

The moles of Cl2 present = Mass of Cl2 / Molecular weight of Cl2 = 3.5 / 71 = 0.049 moles

The moles of PCl5 produced = 2/5 X moles of Cl2 = 2/5 X 0.049 = 0.0196 moles

Mass of PCl5 produced = Moles X molecular weight = 0.0196 moles X 208.2 g / mole = 4.081 grams = 4.1 grams

16)

The balanced equation is

2SO2 + O2 ---> 2SO3

two moles of O2 will react with one mole of O2 to give two moles of SO3

Moles of SO2 present = Mass / Molecular weight = 90 / 64g / mole = 1.41 moles

1.41 moles of SO2 will react with 1.41/2 moles of O2 = 0.705 moles

Moles of O2 present = Mass / Molecular weight = 100 / 32 = 3.125 moles

Moles of O2 left (excess reagent) = 3.125 - 0.705 = 2.42 moles

Mass of O2 left = Moles X molecular weight = 2.42 moles X 32g / mole = 77.44 grams = 77.5grams (approx)

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