Question

A stink bomb creates hydrogen sulfide gas from hydrochloric acid and sodium sulfide solution. The other product is salt. How many liters of hydrogen sulfide gas can be made from 500.0 mL of a 2.00M HCl solution and excess sodium sulfide? The temperature of the reaction is 25.0 °C, and the pressure is 740.0 mmHg.

Answer 1

To calculate the amount of hydrogen sulfide gas produced, we first need to write and balance the chemical equation for the reaction:

HCl(aq) + Na2S(aq) → H2S(g) + 2NaCl(aq)

We can see from the balanced equation that for every 2 moles of HCl reacted, 1 mole of H2S is produced. Therefore, we need to calculate the number of moles of HCl present in the 500.0 mL of 2.00 M solution:

Molarity = moles of solute / liters of solution

2.00 M = moles of HCl / 0.5000 L

moles of HCl = 2.00 M × 0.5000 L = 1.00 moles

Since we have excess sodium sulfide, the limiting reagent is HCl, and we can use the stoichiometry of the balanced equation to calculate the number of moles of H2S produced:

1.00 moles HCl × (1 mole H2S / 2 moles HCl) = 0.500 moles H2S

Now we can use the ideal gas law to calculate the volume of H2S gas produced at 25.0 °C and 740.0 mmHg:

PV = nRT

V = nRT / P

V = (0.500 mol)(0.0821 L·atm/mol·K)(298.2 K) / (740.0 mmHg)(1 atm/760.0 mmHg)

V = 0.0128 L = 12.8 mL

Therefore, 500.0 mL of 2.00 M HCl solution will produce 12.8 mL of H2S gas at 25.0 °C and 740.0 mmHg