Acetylene gas, C2H21g2, can be prepared by the reaction of calcium carbide with water: CaC21s2 + 2 H2O1l2¡Ca1OH221aq2 + C2H21g2 Calculate the volume of C2H2 that is collected over water at 23 °C by reaction of 1.524 g of CaC2 if the total pressure of the gas is 100.4 kPa. (The vapor pressure of water is tabulated in Appendix B.)

Video transcript

Welcome back everyone. We need to consider the reaction of 5.693 g of sodium sulfide. Following the balanced equation below, where solid sodium sulfide reacts with two moles of hydrochloric acid to produce one mole of sodium chloride and one mole of hydrogen sulfide gas. Total pressure of our gas is 121 point to kill a paschal's. We need to calculate the volume of hydrogen sulfide collected over water at Celsius were given the vapor pressure of water at 25 Celsius being 0.313 A T. M. Let's begin by recalling that. We can use our ideal gas equation to find our final answer where we recall that the formula relates pressure of our gas times its volume equal to the molds of our gas times our ideal gas constant R times r temperature in kelvin. However, because we need to find the volume of hydrogen sulfide, we want to isolate for volume so we would divide both sides by pressure so that it cancels out on the left, and we would have that are volume of hydrogen sulfide is equal to the moles of hydrogen sulfide times R gas constant r times r temperature in kelvin, divided by pressure. And so our first two tasks are to find n being our moles of hydrogen sulfide to plug in as well as our pressure of dry hydrogen sulfide since it's collected over water. So let's begin with finding our moles. What we do know from the prompt is our mass, our sample mass of sodium sulfide, which is our first reactant here And we can make note that the molar mass of sodium sulfide from our periodic table is equal to a value of 78.05 g per mole. And so to find our moles of hydrogen sulfide, we want to note down the ratio between hydrogen sulfide and sodium sulfide and we can see from our balanced equation that we have coefficients of one. So this is a 1-1 molar ratio. So these are our coefficients here. And finding our moles of hydrogen sulfide. We would begin with taking our sample mass given in the prompt of 5.693 g of sodium sulfide are reacting and multiplying by our conversion factor to go from grams of sodium sulfide in the denominator, two moles of sodium sulfide in the numerator, In which the denominator will plug in 78.05g equivalent to one mole, which is our molar mass from our periodic table. And then this allows us to cancel out our grams of sodium sulfide and now that we have moles of our reactant. We can multiply by our next piece of geometry, which is to go from molds of our reactant sodium sulfide. Sorry, this is n. a. to us two moles of our gas hydrogen sulfide. And as we stated, we have a 1 to 1 molar ratio and so canceling out moles of sodium sulfide. We're left with molds of our gas hydrogen sulfide and we'll find a total equal to 0.7 to nine moles of our hydrogen sulfide gas. Now, our next step to find our pressure of dry hydrogen sulfide, we want to note the conversion factor that 1 80 M is equivalent to 101.3 25 kg paschal's. And were given our total pressure of hydrogen sulfide equal to 121.2 kg paschal's. We're also given our favorite pressure of water in which our hydrogen sulfide is collected over in A T. M. As 0.313 A. T. M. So we need to convert the total pressure of hydrogen sulfide from killing pascal's two matching units of A T. M. And so multiplying by our conversion factor to go from kila paschal's in the denominator to a T. M's in the numerator. As we stated, we have an equivalent for 1 80 M. Being 101.3 25 Killer paschal's. And so this allows us to cancel out Killer paschal's and we would find that when we take the product here and divide we get a value equal to 1.196 80M as our new total pressure of hydrogen sulfide. But now in atmospheres. And so now that we have this unit in atmospheres. We can go ahead and take the or sorry, we can go ahead and find our pressure of dry hydrogen sulfide by taking our total pressure in a T. M. Subtracted from our vapor pressure of water, which is also given in a t. M. And so below, we would have our total pressure of hydrogen sulfide and A T. M. Sorry. As 11.1 96 A t. M's subtracted from our pressure of water given as 0.313 A t. M. At 25 Celsius. And so this difference in pressure gives us our pressure of dry hydrogen sulfide equal to 1.1 65 A. T. M. Now, with this pressure of hydrogen sulfide, we want to make sure that our temperature is in kelvin because in our ideal gas equation we should have temperature in kelvin so that our units in our gas constant and everything else cancels out appropriately, were given a temperature of 25 Celsius. And so we're going to add to 73.15 to get our kelvin temperature equal to 2 98.15 kelvin. Now we can go into our ideal gas equation where we have, as we stated volume equal to our moles of our gas, which we found for hydrogen sulfide to be 0.0729 moles of hydrogen sulfide. multiplying by our temperature which or sorry by our ideal gas constant, which we should recall is the value 0. units of leaders times A T. M's, divided by moles times kelvin. and then multiplying by our temperature, which we just converted to kelvin as to 98.15 kelvin. Then dividing by our pressure of dry hydrogen sulfide, which we just found as 1.1 65 A t. M. And so now with everything plugged in, notice that we can cancel out our units of a t M. R. Units of moles as well as our units of kelvin, leaving us with leaders as our final unit for volume, which is what we want. And so simplifying this quotient, we would find a value equal to 1.531 liters of hydrogen sulfide collected over water At 25 C. And so for our final answer, what we have highlighted in yellow is our volume of hydrogen sulfide collected over water corresponding to choice D in the multiple choice as our correct answer. So I hope this made sense. And please let us know if you have any questions