H2S + 2NaOH -> Na2S + 2H2O
2.48g of sodium hydroxide is dissolved in water and 1.30g of hydrogen sulfide bubbled through the solution.
a) calculate the mass of NNa2S produced in solution
b) calculate the mass of the excess reagent that remains unreacted
how many moles of HS in 1.30g? and in 2.48g NaOH?
each mole of H2S needs 2 moles of NaOH.
so, ...
2 moles
4 and 6
To answer both parts of the question, we need to use stoichiometry and the concept of limiting reagents. Let's break it down step by step.
Step 1: Write and balance the chemical equation:
H2S + 2NaOH → Na2S + 2H2O
Step 2: Calculate the number of moles of the reactants:
Molar mass of NaOH = 22.99 + 16.00 + 1.01 = 39.99 g/mol
Molar mass of H2S = 1.01 + 32.07 = 33.08 g/mol
Number of moles of NaOH = mass of NaOH / molar mass of NaOH
Number of moles of NaOH = 2.48 g / 39.99 g/mol
Number of moles of H2S = mass of H2S / molar mass of H2S
Number of moles of H2S = 1.30 g / 33.08 g/mol
Step 3: Determine which reactant is the limiting reagent:
To do this, we compare the number of moles of each reactant using the stoichiometry of the balanced equation.
From the balanced equation, we can see that the stoichiometric ratio of NaOH to H2S is 2:1. That means for every 2 moles of NaOH, we need 1 mole of H2S.
So, let's compare the number of moles of NaOH and H2S:
Moles of NaOH / 2 = 2.48 g / 39.99 g/mol / 2 = x mol
Moles of H2S = 1.30 g / 33.08 g/mol = y mol
If x < y, then NaOH is the limiting reagent.
If x > y, then H2S is the limiting reagent.
If x = y, then both reactants are in stoichiometric ratio and none is limiting.
Step 4: Calculate the mass of Na2S produced:
Since NaOH is in excess, we need to calculate the moles of Na2S formed from H2S.
From the balanced equation, the stoichiometric ratio of H2S to Na2S is 1:1. That means 1 mole of H2S reacts to produce 1 mole of Na2S.
So, the moles of Na2S produced = moles of H2S used = y mol
Mass of Na2S produced = moles of Na2S produced x molar mass of Na2S
Step 5: Calculate the mass of the excess reagent that remains unreacted:
Since NaOH is in excess, we need to determine the moles of NaOH remaining after the reaction.
Moles of NaOH remaining = Moles of NaOH initially - moles of Na2S used
Moles of NaOH initially = moles of NaOH calculated in step 2
Mass of NaOH remaining = Moles of NaOH remaining x molar mass of NaOH
I hope this step-by-step explanation helps you solve the problem!