Limestone (CaCO3) reacts with hydrochloric acid according to the ...

Question

Limestone (CaCO3) reacts with hydrochloric acid according to the equation CaCO3 + 2 HCl ---> CaCl2 + H2O + CO2. If 1.00 mol of CO2 has a volume of 22.4 L under the reaction conditions, how many liters of gas can be formed by reaction of 2.35 g of CaCO3 with 2.35 g of HCl? Which reactant is limiting?

Accepted Answer

Hi everyone. This problem reads the equation for the reaction of sodium sulfide with hydrochloric acid is sodium sulfide plus hydrochloric acid yields hydrogen sulfide plus sodium chloride. The volume of one mole of hydrogen sulfide under the reaction conditions is 30.26 liters. How much gas and Leaders is formed by 1.9 g of sodium sulfide and 1.67 g of hydrogen chloride. What is the limiting reactive? So the this question is two parts. The first part we want to answer is how much gas and leaders is formed and then the second is what is the limiting reactant? Okay. And so let's recall that the limiting reactant is completely consumed in a reaction and it control The amount of product formed in order for us to calculate the limiting reactant. We're going to start off with the g of what's given for both of our reactant. Okay so let's go ahead and start with our 1. g of sodium sulfide. Okay, so we want to go from grams of sodium sulfide, two moles of hydrogen sulfide. Okay, so That that is how we're going to figure out how many moles of hydrogen sulfide is produced by this. 1.09 g of sodium sulfide. Okay, so let's go ahead and start off there. So we want to go from grams of sodium sulfide. Two moles of sodium sulfide. So we need the molar mass of sodium sulfide in one mole of sodium sulfide. The molar mass is The Bowler Mass is 78. grams of sodium sulfide. Okay, so making sure our units cancel properly. Our grams of sodium sulfide cancel. And we're left with moles of sodium sulfide. So now we want to go from moles of sodium sulfide, two moles of hydrogen sulfide. And we're going to use the multiple ratio. So let's look at our reaction. Okay our equation says for every one mole of sodium sulfide consumed we have one mole of hydrogen sulfide produced. Okay so our multiple ratio is one. Okay so let's go ahead and write that here for every one mole of hydrogen sulfide. Okay there is one more Of sodium sulfide. So looking at our units here our moles of sodium sulfide cancel. And we're left with moles of hydrogen sulfide. So once we do this calculation will take 1.09, divide by 78.05 and then multiplied by one. This gives us zero 0 to to two moles of hydrogen sulfide produced from that 1.9 g of sodium sulfide. Alright, so now we want to do the same thing but with the amount of hydrochloric acid given. So we have 1.67 g of hydrochloric acid. And our goal is to go from grams of hydrochloric acid, Two moles of hydrogen sulfide. So we want to know how many moles of hydrogen sulfide is going to be produced from this. 1.67 g of hydrogen chloride. Okay so we're going to do the same thing. We want to go from grams of hydrogen chloride, two moles of hydrogen chloride. So we'll use the molar mass. So in one mole of hydrogen chloride The Molar mass is 36. g of hydrogen chloride. Our our grams of hydrogen chloride cancel. And we're left with moles of hydrogen chloride. So now we want to go from moles of hydrogen chloride, two moles of hydrogen sulfide. So we need to look at the multiple ratio here. Okay, so we C in our equation that for every one mole of hydrogen chloride consumed, there is one mole of hydrogen sulfide produced are multiple ratio is 1-1. So let's go ahead and write that in. Okay, so making sure our units cancel here are moles of hydrogen chloride cancel. And we're left with moles of hydrogen sulfide. So we'll do this calculation and we get 0. 80 moles of hydrogen sulfide. So we need to determine which one is the limiting reactant. The one that's the limiting reactant is going to be the one that is the lesser value. So we see from both of the givens how much moles of hydrogen sulfide is going to be produced. So the one that is in the lesser value is the limiting reactant. This is going to be the one that's completely consumed in the reaction and controls the amount of product formed. So that makes our our sodium sulfide, the limiting reactant. Okay, okay, so our sodium sulfide is the limiting reactant. Alright, so now the question asked us how much gas and leaders is formed. Okay, so we want to go from Moles. So we know our limiting reactant, we know that it is 0. moles of hydrogen sulfide. So now what our goal is to do is we want to go from moles of hydrogen sulfide to leaders. Okay, and the way that we can do this is it's given in the problem that the volume of one mole of hydrogen sulfide under the reaction conditions is 30.26 leaders. So we're going to going to go ahead and use that as our conversion. Okay, so in one mole of hydrogen sulfide there is 30.26 liters. Alright, so our moles of hydrogen sulfide cancel and we're left with volume. Okay, so once we do this calculation, we get an answer of 0. liters. Okay, so in terms of our final answer, we're going to go ahead and write that sodium sulfide is the limiting reactant. Okay, And the volume of gas is 0.770. L of gas formed. Alright, so let's go ahead and highlight that these are our final answers and that is the end of this problem. I hope this was helpful

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