When hydrogen sulfide gas is bubbled into a solution of sodium hydroxide, the reaction forms sodium sulfide and water. How many grams of sodium sulfide are formed of 2.68 g. of hydrogen sulfide is bubbled into a solution containing 0.0823 g. of sodium hydroxide, assuming that the limiting reagent is completely consumed

0.0803g of Na2S are produced

Explanation:

Hydrogen sulfide, H2S, reacts with sodium hydroxide, NaOH, to produce sodium sulfide, Na2S and water.

2NaOH + H2S → 2H2O + Na2S

To solve this question we must find the moles of each reactant. With the chemical quation we can find the limiting reactant. With limiting reactant we can find the moles of Na2S and its mass

Moles NaOH -Molar mass: 40g/mol-

0.0823g * (1mol / 40g) = 0.002058moles

Moles H2S -Molar mass: 34.082g/mol-

2.68g * (1mol / 34.082g) = 0.0786 moles

For a complete reaction of 0.0786 moles of H2S are needed:

0.0786 moles H2S * (2mol NaOH / 1mol H2S) = 0.157 moles NaOH

As there are just 0.002058 moles, NaOH is limiting reactant.

The moles of Na2S produced are:

0.002058moles NaOH * (1mol Na2S / 2mol NaOH) = 0.001029 moles Na2S

Mass Na2S -Molar mass: 78.0452g/mol-

0.001029 moles Na2S * (78.0452g / mol) =