How many milliliters of 0.150 M BaCl₂ are needed to...

Question

How many milliliters of 0.150 M BaCl₂ are needed to react completely with 35.0 mL of 0.200 M Na₂SO₄? How many grams of BaSO₄ will be formed?

Accepted Answer

Welcome back, everyone. We're told that the lead two plus ion can be precipitated using our chloride an ion, calculate the volume in milliliters of 20.11 molar hydrochloric acid solution required to react with 40 mL of 400.26. Mohler led to nitrate determined the mass in grams of lead chloride, which is formed. Our first step is to write out our reaction in which we're going to show our lead to nitrate PB parentheses. N 03, sub two, the solution of lead to nitrate is reacting with hydrochloric acid HCL, which we should recall is one of our memorized strong assets from, from general chemistry and as products we're going to form solid lead chloride as an insoluble precipitate. And then we would form nitric acid H N 03 as our second product. Now, we need to make sure that this is balanced. So we're going to place a coefficient of two in front of our HCL and two in front of our nitric acid in order to have a balanced equation. Now, with this bounced equation, let's move into part one of our prompt, which is to calculate our volume in milliliters of hydrochloric acid that is required to react with our lead to nitrate. Now note that we do not know our moles of hydrochloric acid. So we're going to need to solve that first before getting to our volume. And in order to figure this out, we can interpret that our molar itty or molar concentration is expressed in moles per liter where we can take our given 0.26 Mohler of lead to nitrate and interpret it as 0.26 moles of lead to nitrate per leader of solution where we can multiply to cancel out our unit of leaders by the next conversion factor where we would recall that our prefix Milli tells us that we have in our denominator for one male leader in our numerator and equivalent of 10 to the negative third power leaders. So canceling out leaders were left with moles per mil leader where we can multiply by our volume of lead to nitrate given as point oh male leaders in our prompt of PBN 032. So now canceling out our unit of male leaders, we just have moles, but we need moles of HCL. So we're going to multiply by the molar ratio where in our denominator, we have moles of lead to nitrate and in our numerator, we have moles of HCL. And so looking at our balanced equation, we have in our denominator, one mole of lead to nitrate for two moles of hydrochloric acid So canceling out moles of lead to nitrate. We're left with moles of hydrochloric acid as our final unit. So we'll move the this down our volume of male leaders of hydrochloric acid. And for our moles of hydrochloric acid. Once we take the product of each of our quotients, here, we're going to get a result of 0. moles of hydrochloric acid HCL. So now with this moles of hydrochloric acid, we can find our volume of hydrochloric acid required to react where we would begin with our moles, which we just found as .0208 moles of HCL and given by the molar concentration of hydrochloric acid in our prompt, we can multiply this by or as a conversion factor where we would interpret that we have 0.11 moles of HCL and denominator per mole of solution or sorry, per leaders of solution to be accurate. Because again, molar concentration is in terms of moles per liter as we outlined above. So we can now cancel out our units of moles of HCL. We have leaders of HCL, but we need our unit of volume to be in middle leaders. So we're going to multiply by the next conversion factor where we recall, as we stated before, one millimeter in our numerator has an equivalence of 10 to the negative third power of our base unit leader in the denominator. And so now canceling out leaders were left with ml of hydrochloric acid as our final unit. And in our calculators, this will result in a volume of 189. milliliters of hydrochloric acid that is required to react with our lead to nitrate. And we can round this to about three sig figs as mL of HCL. So this would be our first answer so far. Now, we just need our next answer, which is our mass and grams of solid lead chloride, which is formed, we're sorry, rather lead to chloride. So let's go ahead and figure that out below for our mass of lead to chloride. So we're just going to utilize the geometry beginning with our given moles of our reactant led to nitrate were given . moles of lead to nitrate per liter of solution. So we're just reinterpreting its given Mueller concentration. Again, we need to cancel out our unit of volume and leaders. So we're going to multiply by the conversion factor where 10 to the negative third power leaders in the numerator is equivalent to one millimeter in the denominator. Canceling out leaders were now left with moles per mil leader and we need to multiply by the volume of lead to nitrate our reactant, which is used given us 40 mL according to our prompts. So now canceling out our militar units were left with moles of lead to nitrate where we need to get two moles of our lead to chloride product. So in our denominator, we're going to multiply by the conversion factor where we have moles of lead to nitrate in the denominator. And then we have moles of lead to chloride in the numerator. So, utilizing our molar ratio from our balanced equation, we see that we have for one mole of lead to nitrate in the denominator and equivalence of two moles of or sorry, one mole of lead to chloride, which is produced on the product side. So we have a 1-1 molar ratio, canceling out our moles of our lead to nitrate reactant. We're left with our final unit which has led to chloride moles. And actually, we need to keep going because we need again mass of lead to chloride. So we're going to multiply by our next conversion factor. Where in our denominator, we need moles of lead to chloride P BCL two. And in our numerator, we need grams of lead to chloride. So let's recall from our periodic tables, our molar mass which tells us that we have 278.1 g of lead to chloride per mole of lead to chloride, canceling out moles of lead to chloride. We finally have the correct unit being grams of lead to chloride as our final unit. And in our calculators, the product of these questions will result in a value of 2.892 to 4 g of lead to chloride, which is produced. And we can also round this 2366 as 2.89 g of lead to chloride again produced or formed. And so this would be our second final answer. So our two final answers will correspond to choice D and the multiple choice to complete this example. I hope that everything I went through made sense. Let us know if you have any questions and I'll see everyone in the next video.

How many milliliters of 0.150 M BaCl₂ are needed to react completely with 35.0 mL of 0.200 M Na₂SO₄? How many grams of BaSO₄ will be formed?

Key Concepts

Stoichiometry

Molarity

Precipitation Reaction

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Key Concepts

Stoichiometry

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How many milliliters of 0.150 M BaCl₂ are needed to react completely with 35.0 mL of 0.200 M Na₂SO₄? How many grams of BaSO₄ will be formed?