Question

How many grams of sodium sulfide are formed if 1.90 g of hydrogen sulfide is bubbled into a solution containing 2.53 g of sodium hydroxide, assuming that the sodium sulfide is made in 94.0% yield?

A) 1.45 g
B) 2.07 g
C) 2.39 g
D) 2.55 g

Answer 1

To find the number of grams of sodium sulfide formed, we need to determine the limiting reactant. Based on the stoichiometry and given masses, hydrogen sulfide is the limiting reactant. Using the 94.0% yield, the amount of sodium sulfide formed is 4.35 g. The correct answer is A) 1.45 g.

Step-by-step explanation:

To find the number of grams of sodium sulfide formed, we first need to determine the limiting reactant in the reaction between hydrogen sulfide (H2S) and sodium hydroxide (NaOH). The balanced equation for the reaction is:

H2S + 2NaOH → Na2S + 2H2O

Now, let's calculate the number of moles of H2S and NaOH. First, we convert the given masses of H2S and NaOH to moles using the molar masses:

H2S: 1.90 g / (32.07 g/mol) = 0.0593 mol

NaOH: 2.53 g / (39.997 g/mol) = 0.0633 mol

Based on the stoichiometry of the reaction, H2S and NaOH have a 1:2 mole ratio. Therefore, the limiting reactant is H2S because there is less moles of H2S compared to NaOH.

Next, we can calculate the moles of Na2S formed using the 94.0% yield:

0.0593 mol H2S × 1 mol Na2S / 1 mol H2S × 0.94 = 0.0558 mol

Finally, we convert the moles of Na2S to grams using the molar mass:

0.0558 mol Na2S × (78.045 g/mol) = 4.35 g

Therefore, the answer is A) 1.45 g.