H2
S + HNO3 --------> NO + H2
O + S
150 g of hydrogen sulfide are dissolved in 3500 mL of water.
a. What is the molarity of the hydrogen sulfide solution?
b. How many grams of sulfur can be produced from 50 mL of
this solution?
c. 22.4 L of nitrogen monoxide gas are produced at STP. What
volume of the solution was used?
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To answer these questions, we'll need to use stoichiometry and molarity calculations.
a. What is the molarity of the hydrogen sulfide solution?
To find the molarity (M) of a solution, we need to know the moles of solute (hydrogen sulfide) and the volume of the solution.
First, let's determine the number of moles of hydrogen sulfide. We are given that 150 g of hydrogen sulfide is dissolved in 3500 mL of water. The molar mass of hydrogen sulfide (H2S) is 2 g/mol + 32 g/mol = 34 g/mol.
Number of moles of H2S = mass of H2S / molar mass of H2S
= 150 g / 34 g/mol
= 4.41 mol
Next, we need to find the volume of the solution in liters (L) since molarity is expressed in moles of solute per liter of solution.
Volume of solution = 3500 mL = 3500 mL * (1 L / 1000 mL)
= 3.5 L
Now, we can calculate the molarity of the H2S solution.
Molarity (M) = moles of H2S / volume of solution (in L)
= 4.41 mol / 3.5 L
= 1.26 M
Therefore, the molarity of the hydrogen sulfide solution is 1.26 M.
b. How many grams of sulfur can be produced from 50 mL of this solution?
From the balanced chemical equation, we can see that 1 mole of H2S produces 1 mole of S.
So, if we have the moles of H2S, we can directly convert it to moles of sulfur.
Number of moles of S = moles of H2S
We already calculated the moles of H2S in part (a) as 4.41 mol.
So, the number of moles of sulfur produced = 4.41 mol
Next, we need to calculate the mass of sulfur using its molar mass. The molar mass of sulfur (S) is 32 g/mol.
Mass of sulfur = moles of sulfur * molar mass of sulfur
= 4.41 mol * 32 g/mol
= 141 g
Therefore, 50 mL of the solution can produce 141 grams of sulfur.
c. 22.4 L of nitrogen monoxide gas are produced at STP. What volume of the solution was used?
Since we are given the volume and assuming the reaction occurred at STP (standard temperature and pressure), we can use the ideal gas law.
According to the balanced chemical equation, 1 mole of H2S produces 1 mole of NO.
Number of moles of NO = 22.4 mol (given)
From the stoichiometry, we can determine the number of moles of H2S.
Number of moles of H2S = Number of moles of NO
Using the molarity of the solution from part (a) (1.26 M), we can calculate the volume of the solution (in liters) used to produce 22.4 L of NO.
Volume of solution (in L) = Number of moles of H2S / Molarity of H2S
= 22.4 mol / 1.26 M
= 17.78 L
Therefore, the volume of the solution used to produce 22.4 L of nitrogen monoxide gas (NO) at STP is 17.78 L.