Consider the following balanced chemical equation: H 2 O+ 2 MnO 4 – + 3 SO 3 2- → 2 MnO 2 + 3 SO 4 2- + 2 OH – How many grams of MnO 2 (MW:86.94 g/mol) will be created when 25.0 mL of 0.120 M MnO 4 – (MW:118.90 g/mol) reacts with 32.0 mL of 0.140 M SO 3 2- (MW:80.07 g/mol).

In this practice question it says, consider the following balanced chemical equation. They say how many grams of Manganese 4 oxide, which has a molecular weight of 86.94 grams per mole, we created when twenty 5 ml of 0.120 molar permanganate ion reacts with 32 milliliters of 0.140 molar sulfide ion. Alright. So they want us to find the grams of Manganese 4 Oxide. So this is our grams of unknown. And what they're giving us here is volume of molarity. Again, we've seen this numerous times. Liters times molarity will give us moles. So they're giving us, technically here, moles of given. And here, they're technically giving us here moles of given. So we have more than one reactant with a given amount, which means that we're gonna have to do stoichiometry twice and then compare to see how much each reactant says it can make in terms of our product. Out of those choices, we'll decide which one is our correct answer. Alright. So let's just start out with the permanganate ion. We're gonna say 25 ml's. We're gonna convert those ml's into liters. So one ml is 10 to the negative 3 liters. Remember that molarity is a converting factor, So that's point 120 moles of manganese a permanganate ion per 1 liter. Now that we have moles of given officially, we can do a mole to mole comparison to see how many moles of Manganese 4 oxide will be created. So according to our balanced equation, for every 2 moles of this, there are 2 moles of this. So we have 2 moles of permanganate on the bottom and then 2 moles of manganese 4 oxide on top. Then finally, we're looking for grams of manganese 4 oxide. So we're gonna say for every 1 mole of it, we're told that its weight is 86.94 grams. So here moles cancel out and we're gonna have as our answer at so far as point 8 as 0.260-829 grams, Manganese 4 oxide. Now, we're gonna do the same thing with sulfide ion. So we're gonna start out with 32 ml's, convert those ml into liters first, so 1 milli is 10 to the negative 3. Then we're gonna say remember molarity is a conversion factor, so we have 0.140 moles of sulfide ion per 1 liter. Now that we have the moles of Gibbon, we do a mole to mole comparison to get to the moles of manganese 4 oxide. K. For every 2 moles of Manganese 4 oxide. And then finally, we're gonna convert. We're gonna say 1 mole of manganese 4 oxide weighs 86.94 grams. So when we do that we get as our answer 0.259660 8 grams of Manganese 4 Oxide. So those numbers are really close together. But remember, the smaller answer is the correct choice, that represents our theoretical yield. So this number is slightly smaller, so this is our theoretical yield. The reactant that says it can make that smaller answer represents our limiting reactant. So sulfide ion is our limiting reactant. I know the question doesn't ask for it, but we can say it. So limiting reactant or limiting reagent, which would mean permanganate ion is the excess reactant or excess reagent. Here, if we want sig figs we're gonna say that this number here has 3 sig figs, so does this one, so does this one, and so does this one. So in terms of sig figs, write down our answer as 0.260 grams of permanganate if we wanted the correct number of significant figures. So just realize here that this is a little bit different, in terms of other questions you might have seen at this point. But remember, they're giving you multiple given amounts. You have to do stoichiometry multiple times, compare your answers to one another. The smaller answer is the correct choice. It represents your theoretical yield.

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6. Chemical Quantities & Aqueous Reactions

Solution Stoichiometry

General Chemistry

6. Chemical Quantities & Aqueous Reactions

Solution Stoichiometry

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Multiple Choice

Consider the following balanced chemical equation:
H₂O+ 2 MnO₄^– + 3 SO₃^2- → 2 MnO₂ + 3 SO₄^2-+ 2 OH^–
How many grams of MnO₂ (MW:86.94 g/mol) will be created when 25.0 mL of 0.120 M MnO₄^– (MW:118.90 g/mol) reacts with 32.0 mL of 0.140 M SO₃^2- (MW:80.07 g/mol).

0.073 g

0.089 g

0.19 g

0.26 g

0.67 g

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Verified step by step guidance

Step 1: Begin by identifying the limiting reactant. Calculate the moles of MnO4– and SO32– using their respective concentrations and volumes. Use the formula: \( \text{moles} = \text{concentration} \times \text{volume} \).

Step 2: For MnO4–, use the concentration (0.120 M) and volume (25.0 mL). Convert the volume from mL to L by dividing by 1000. Calculate the moles of MnO4–.

Step 3: For SO32–, use the concentration (0.140 M) and volume (32.0 mL). Convert the volume from mL to L by dividing by 1000. Calculate the moles of SO32–.

Step 4: Use the stoichiometry of the balanced equation to determine which reactant is the limiting reactant. The balanced equation shows that 2 moles of MnO4– react with 3 moles of SO32–. Compare the mole ratio of the reactants to find the limiting reactant.

Step 5: Once the limiting reactant is identified, use its moles to calculate the moles of MnO2 produced using the stoichiometry of the balanced equation. Then, convert moles of MnO2 to grams using its molar mass (86.94 g/mol) with the formula: \( \text{mass} = \text{moles} \times \text{molar mass} \).