Nickel(II) sulfate, used for nickel plating, is prepared by treat-ment of nickel(II) carbonate with sulfuric acid: NiCO 3 + H 2 SO 4 → NiSO 4 + CO 2 + H 2 O (b) How many grams of NiSO4 are obtained if the yield is 78.9%?

Hello everyone today, we are being given the following question and asked to solve for the mass of selenium oxide. So we have the following reaction that has a yield of 96%. We have a balanced chemical equation here We are being asked to determine the mass of selenium oxide produced if 20 g of hydrogen selenium And 40 g of oxygen gas were reacted together. So the first thing I want to do is find the limiting reactant. And to do that we're gonna take each Amount of reaction that we have and convert it into most of a product. And in this case we're going to the product of selenium oxide. So we're gonna take our 20 g of H two s. c. And to convert it to moles of our product. We first want to do the molar mass. So we're gonna say one mole of H two S. E. Is equal to its molar mass, which is 80.98 grounds H two s. E. And then since we want to get moles of our products, lithium oxide, we have to do the multiple ratio and we do that by using the coefficients that are in front of each reactant or product. So we have two moles of H two S. E. And we put that in the denominator so that our units can cancel later on. And for every two moles of H two S. C. We produce two moles of S. E. 02 or selenium oxide. Our units cancel as such. And we're left with a value of 0. malls of selenium oxide. We're gonna do the same process with the oxygen gas. We're gonna take our 40 g of 02. We're gonna do our mole mass. We're going to say one mole of 02 Is 32 g. And then we're gonna do the multiple ratio once again. But instead we're going to say we have three moles of 02. Right? Using that coefficient and that's going to produce two moles of psyllium oxide. Our units cancel once again and we are left with 0.833 moles of sodium oxide. Now limiting reaction is the reaction that produces the least amount of product. And in that case that is going to be our H two S. E. So we're gonna make a note here, H two SC. Is our limiting reactant. Our 2nd step is using that Mole using the moles of our limited reactant and converting that into grams of our product. So we're going to say 0.2470 moles. We're going to use the molar mass for this, we're going to say one mole of S. C. 02 Is equal to the Molar mass which is 110.96 g. Our units cancel. And we're left with a value of 27.40 g of selenium oxide. Our third and final step is going to be solving for the actual percent yield that is going to be the actual yield Over times the theoretical yield, I was gonna say theoretical, That gives us our actual which is 96%. So they're going to say 96 over 100 times our theoretical, which is 27.40 grams of selenium oxide, Which when we do the math yields us 26. grams of selenium oxide. As our final answer, I hope this helped, and until next time.

Ch.3 - Mass Relationships in Chemical Reactions

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McMurry 8th Edition

Ch.3 - Mass Relationships in Chemical Reactions

Problem 72b

Chapter 3, Problem 72b

Nickel(II) sulfate, used for nickel plating, is prepared by treat-ment of nickel(II) carbonate with sulfuric acid: NiCO₃ + H₂SO₄ → NiSO₄ + CO₂ + H₂O
(b) How many grams of NiSO4 are obtained if the yield is 78.9%?

Verified step by step guidance

Identify the balanced chemical equation: \( \text{NiCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{NiSO}_4 + \text{CO}_2 + \text{H}_2\text{O} \).

Determine the molar mass of each compound involved: \( \text{NiCO}_3 \), \( \text{H}_2\text{SO}_4 \), and \( \text{NiSO}_4 \).

Calculate the theoretical yield of \( \text{NiSO}_4 \) using stoichiometry based on the initial amount of \( \text{NiCO}_3 \) or \( \text{H}_2\text{SO}_4 \) provided.

Apply the percentage yield formula: \( \text{Actual Yield} = \text{Theoretical Yield} \times \frac{\text{Percentage Yield}}{100} \).

Convert the actual yield from moles to grams using the molar mass of \( \text{NiSO}_4 \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions based on the balanced chemical equation. It allows chemists to determine the amount of product formed from a given amount of reactant. In this case, understanding the stoichiometric ratios from the reaction between nickel(II) carbonate and sulfuric acid is essential to calculate the theoretical yield of nickel(II) sulfate.

Percent Yield

Percent yield is a measure of the efficiency of a chemical reaction, calculated by comparing the actual yield of a product to the theoretical yield. It is expressed as a percentage and is crucial for evaluating how much of the expected product was actually produced. In this question, the yield of 78.9% indicates that only a portion of the theoretical amount of NiSO4 was obtained.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is essential for converting between the number of moles of a substance and its mass. To find the mass of NiSO4 produced, one must first calculate its molar mass and then apply it to the stoichiometric calculations involving the yield.

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Nickel(II) sulfate, used for nickel plating, is prepared by treat-ment of nickel(II) carbonate with sulfuric acid: NiCO3 + H2SO4 → NiSO4 + CO2 + H2O (b) How many grams of NiSO4 are obtained if the yield is 78.9%?

Verified video answer for a similar problem:

Key Concepts

Stoichiometry

Percent Yield

Molar Mass

Nickel(II) sulfate, used for nickel plating, is prepared by treat-ment of nickel(II) carbonate with sulfuric acid: NiCO₃ + H₂SO₄ → NiSO₄ + CO₂ + H₂O
(b) How many grams of NiSO4 are obtained if the yield is 78.9%?