How many grams of sodium sulfide are formed if 1.25 g of hydrogen sulfide is bubbled into a solution containing 2.00 g of sodium hydroxide, assuming that the sodium sulfide is made in 92.0 % yield?

Question

 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 if 1.25 g of hydrogen sulfide is bubbled into a solution containing 2.00 g of sodium hydroxide, assuming that the sodium sulfide is made in 92.0 % yield?

Answer 1

Hi, Sandra,

We need to have a balanced equation. Here's what I came up with:

Next, we need to determine the number of moles we have of each of the two starting reagents. Determine the molar mass of both and then convert the grams of each into moles of each.

For H2S, the molar mass is 34.1 g/mole. The moles of H2S is obtained by taking the grams (1.25 g) and dividing by its molar mass. 1.25g/(34.1g/mole) = 0.037 moles. Note how the units cancel in this division. Grams cancels and moles moves to the top.

Do the same exercise for NaOH. I get 0.050 moles.

Our equation tells us we need twice as many mole of NaOH as of H2S. We don't have enough NaOH to react with all the H2S. So the reaction is limited by the NaOH, so we'll use that material to determine how much sodium sulfide is formed.

0.050 moles of NaOH should produce 0.025 moles of Na2S, since our balanced equation says we only get 1 mole of Na2S for every 2 moles of NaOH.

Convert that to grams using the molar mass for Na2S: 0.025 mole * 78.1g/mole = 1.95 grams (3 sig figs)

I'm not clear why we are ask to assume a 92.0% yield, unless they assume my old lab partner was involved. So multiply 0.92 times the theoretical mass of 1.95 grams to find the answer.

I hope this helps,

Bob

Answer 2

Write a correctly balanced equation for the reaction taking place:

H₂S(g) + 2NaOH(aq) ==> Na₂S(aq) + 2H₂O(l) ... balanced equation

Find limiting reactant:

moles H2S present: 1.25 g H2S x 1 mol H2S/34.1 g = 0.03666 moles

moles NaOH present = 2.00 g NaOH x 1 mol NaOH/40 g = 0.0500 moles

NaOH is limiting because according to the balanced equation we need TWO moles NaOH for each mole of H2S, and we don't have that amount.

Theoretical yield:

0.0500 mol NaOH x 1 mol Na₂S/2 mol NaOH x 78 g Na₂S/mol = 1.95 g Na₂S

This represents 100% yield.

If the yield is only 92.0%, then the mass obtained = 0.920 x 1.95 g = 1.78 g