Question

4.8

Sodium sulfide reacts with sulfuric acid to produce sodium sulfate and hydrogen sulfide. Assume that excess sulfuric acid is allowed to react with 10.0 g of sodium sulfide and calculate:

a) the moles of sodium sulfide used.

b) the moles of hydrogen sulfide liberated.

c) the grams of hydrogen sulfide liberated.

d) the volume of hydrogen sulfide liberated at STP.

Answer: a)0.128 mole; b) 0.128 mole; c) 4.36 g; d) 2.87 liters

Answer 1

Na2S + H2SO4 ------------------ Na2SO4 + H2S

1 mole                                      1 mole        1 mole

mass of Na2S = 10.0g

molar mass of Na2S = 78.05 g/mole

number of moles of Na2S = mass/molar mass

number of moles of Na2S = 10.0/78.05=0.128 moles

a) number of moles of Na2S used = 0.128 moles

b)

According to equation

1mole of Na2S = 1 mole of H2S

0.128 mole of Na2S=?

                                 = 0.128x1/1= 0.128 moles of H2S

number of moles of H2S liberated = 0.128 moles

c) number of moles of H2S liberated = 0.128 moles

molar mass of H2S= 34.1 g/mole

mass of one mole of H2S= 34.1 g

mass of 0.128 moles of H2S= 0.128x34.1= 4.3648 grams

mass of H2S produced = 4.36 grams

d)

Volume of one mole of H2S at STP= 22.4L

Volume of 0.128 mole of H2S = ?

                                                = 0.128x22.4/1= 2.8672 L

Volume of H2S liberated = 2.87 Liters.