Answer:
So, approximately 2.48 grams of sodium sulfide (Na2S) are formed when 2.5 grams of hydrogen sulfide (H2S) is bubbled into a solution containing 1.85 grams of sodium hydroxide (NaOH), assuming the limiting reagent is completely consumed.
Step-by-step explanation:
To determine the grams of sodium sulfide formed, we first need to identify the limiting reagent in the reaction. The limiting reagent is the reactant that will be entirely consumed, thus determining the maximum amount of product that can be formed.
The balanced chemical equation for the reaction is:
H2S (hydrogen sulfide) + 2 NaOH (sodium hydroxide) → Na2S (sodium sulfide) + 2 H2O (water)
Now, let's calculate the number of moles for each reactant:
1. Moles of H2S:
Given mass of H2S = 2.5 g
Molar mass of H2S = 1.00794 g/mol (H) + 32.065 g/mol (S) = 33.07394 g/mol
Number of moles of H2S = 2.5 g / 33.07394 g/mol ≈ 0.0757 mol
2. Moles of NaOH:
Given mass of NaOH = 1.85 g
Molar mass of NaOH = 22.98977 g/mol (Na) + 15.9994 g/mol (O) + 1.00794 g/mol (H) = 40.99711 g/mol
Number of moles of NaOH = 1.85 g / 40.99711 g/mol ≈ 0.0451 mol
Now, to find the limiting reagent, we compare the moles of each reactant to their stoichiometric coefficients in the balanced equation. The stoichiometric coefficient of H2S is 1, and the stoichiometric coefficient of NaOH is 2.
The molar ratio of H2S to NaOH is:
0.0757 mol H2S / 0.0451 mol NaOH ≈ 1.679
Since the ratio is greater than 1, it means we have more moles of H2S than NaOH. Therefore, NaOH is the limiting reagent.
Now, we calculate the moles of Na2S formed from the limiting reagent, NaOH.
From the balanced equation, the stoichiometric ratio of NaOH to Na2S is 2:1.
Moles of Na2S formed = 0.0451 mol NaOH (since NaOH is the limiting reagent)
Finally, we find the mass of Na2S formed:
Molar mass of Na2S = 22.98977 g/mol (Na) + 32.065 g/mol (S) = 55.05477 g/mol
Mass of Na2S formed = Moles of Na2S formed × Molar mass of Na2S
Mass of Na2S formed = 0.0451 mol × 55.05477 g/mol ≈ 2.48 g