How many milliliters of barium oral suspensions containing 250 mg and 500 mg per 5 mL should be

How many milliliters of barium oral suspensions containing...

How many milliliters of barium oral suspensions containing $250 \mathrm{mg}$ and $500 \mathrm{mg}$ per $5 \mathrm{~mL}$ should be used in preparing $500 \mathrm{~mL}$ of a suspension containing $55 \mathrm{mg}$ of barium per $\mathrm{mL}$ ?

Key Concepts

Concentration Calculations

This concept involves understanding how much of the active substance is present in a given volume of solution. It is critical for determining how different concentrate levels contribute to a desired overall concentration when mixed.

Mixture Problems

Mixture problems require combining substances or solutions of different concentrations to obtain a final mixture with a specific concentration. Solving these problems involves balancing the amount of the active ingredient contributed by each input relative to the total volume.

Systems of Linear Equations

This concept is used to set up relationships between various components in a mixture problem. By establishing equations for the total amount of solute and the total volume of the mixture, one can solve for unknown quantities using methods for solving simultaneous linear equations.

Proportions and Ratios

Proportions involve the relationship between parts and the whole, which is especially useful in mixing different concentrations. By using ratios, one can determine the correct amounts of each component needed to achieve the desired final concentration.

Transcript

00:01 Chapter 6, problem 60, asks us for the mass of barium sulfite solid, produced when reacting a given amount of barium chloride and iron 3 sulfate.

00:12 So, to start with this problem, we should first write a balanced chemical equation to help us see what's happening.

00:19 So we're told that we're starting with barium chloride and iron 3 sulfate.

00:26 So we'll start by writing those two reactants.

00:28 So barium chloride plus iron sulfate.

00:40 Next, we can put in the products by switching the ions here.

00:44 So that gives us b .a .s .4, and that's our solid precipitate here.

00:51 That's the barium sulfate we're asked about.

00:55 And iron chloride.

00:58 Next, let's make sure this equation is balanced.

01:02 So going across, we see that barium has one on each side, though that's already balanced.

01:07 But chlorine is not balanced here.

01:10 We have two on the left, but three on the right.

01:13 So if we multiply the left by three and the right by two, we'll get six chlorines on each side, which is balanced.

01:21 So that puts a three in front of bacl2 and a two in front of fecl3.

01:28 Now our balance of barium has changed, however.

01:31 We now have three on the left, but only one on the right.

01:33 So to balance that, we'll add a three in front of baso4.

01:40 Going across, we now see that everything is balanced.

01:44 So now we can start solving the problem.

01:46 So we're told that we have 100 mils of 0 .1 molar solution for each of our two reactants.

01:54 So that's 100 mils of 0 .1 molar for each of the two.

02:04 So our process here will be to convert each of these to moles to figure out how many moles of each reactant we have...

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