An alternative method for producing hydriodic acid is the reaction of iodine with hydrogen sulfide:H 2 S + I 2 → 2 HI + S (b) How many grams of HI are produced from the reaction of 95.4 g of H 2 S with excess I 2 ?

Hey everyone. Our question here is asking us to calculate the g of product that form 123.59 g of oxygen gas completely reacts. Now, looking at our chemical reaction, we can see that it is completely balanced so we don't need to do any other work on that. And essentially, our question is asking us to go from grams of oxygen into moles of oxygen and relating that to moles of strontium oxide and then finally into grams of strontium oxide. Now that we have our pathway, let's go ahead and begin with our 23.59 g of oxygen gas. Now, in order to convert this into moles, we're going to need the molar mass of oxygen gas. So the molar mass of oxygen gas is going to be 16 times two, which is 32. In putting that into our dimensional analysis, we know that we have 32.0 g per one mole. And now looking at our reaction, we can see that per one mole of oxygen. We have two moles of strontium oxide and in order to convert this into grams of strontium oxide, we're going to need to calculate the molar mass of strontium oxide. And when we do that, we find that it's 103 .62 g per mole. Plugging that into our dimensional analysis. We know that per one mole of strontium oxide, we have 103.62 g of strontium oxide. And when we calculate this out, we end up with a value of .77 g of strontium oxide. And since we have four significant figures, We can round this out to 152. and this is going to be our final answer. So I hope this made sense and let us know if you have any questions.

Ch.3 - Mass Relationships in Chemical Reactions

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McMurry 8th Edition

Ch.3 - Mass Relationships in Chemical Reactions

Problem 71b

Chapter 3, Problem 71b

An alternative method for producing hydriodic acid is the reaction of iodine with hydrogen sulfide:H₂S + I₂ → 2 HI + S (b) How many grams of HI are produced from the reaction of 95.4 g of H₂S with excess I₂?

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Identify the balanced chemical equation: \( \text{H}_2\text{S} + \text{I}_2 \rightarrow 2 \text{HI} + \text{S} \).

Calculate the molar mass of \( \text{H}_2\text{S} \) using the atomic masses of hydrogen and sulfur.

Determine the number of moles of \( \text{H}_2\text{S} \) by dividing the given mass (95.4 g) by its molar mass.

Use the stoichiometry of the balanced equation to find the moles of \( \text{HI} \) produced. According to the equation, 1 mole of \( \text{H}_2\text{S} \) produces 2 moles of \( \text{HI} \).

Calculate the mass of \( \text{HI} \) produced by multiplying the moles of \( \text{HI} \) by its molar mass.

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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 substances consumed and produced in a reaction. In this case, understanding the stoichiometric coefficients from the balanced equation is essential to calculate how many grams of HI can be produced from a given mass of H2S.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is crucial for converting between grams and moles, which is necessary for stoichiometric calculations. For this problem, calculating the molar mass of H2S and HI will enable the conversion of the mass of H2S into moles, facilitating the determination of the mass of HI produced.

Limiting Reactant

The limiting reactant is the substance that is completely consumed first in a chemical reaction, thus determining the maximum amount of product that can be formed. In this scenario, since I2 is in excess, H2S is the limiting reactant. Identifying the limiting reactant is essential for accurately calculating the yield of HI produced from the reaction.

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Aluminum reacts with oxygen to yield aluminum oxide. If 5.0 g of Al reacts with 4.45 g of O2, what is the empirical formula of aluminum oxide?

Textbook Question

The industrial production of hydriodic acid takes place by treatment of iodine with hydrazine (N2H4): 2 I2 + N2H4 --> 4 HI + N2 (b) How many grams of HI are produced from the reaction of 115.7 g of N2H4 with excess iodine

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Textbook Question

An alternative method for producing hydriodic acid is the reaction of iodine with hydrogen sulfide: H₂S + I₂ → 2 HI + S (a) How many grams of I₂ are needed to react with 49.2 g of H₂S?

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Textbook Question

Nickel(II) sulfate, used for nickel plating, is prepared by treat-ment of nickel(II) carbonate with sulfuric acid: NiCO₃ + H₂SO₄ → NiSO₄ + CO₂ + H₂O (a) How many grams of H₂SO₄ are needed to react with 14.5 g of NiCO₃?

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Textbook Question

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%?

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Textbook Question

Hydrazine, N₂H₄, once used as a rocket propellant, reacts with oxygen: N₂H₄ + O₂ → N₂ + 2 H₂O (a) How many grams of O2 are needed to react with 50.0 g of N₂H₄?

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An alternative method for producing hydriodic acid is the reaction of iodine with hydrogen sulfide:H2S + I2 → 2 HI + S (b) How many grams of HI are produced from the reaction of 95.4 g of H2S with excess I2?

Verified video answer for a similar problem:

Key Concepts

Stoichiometry

Molar Mass

Limiting Reactant

An alternative method for producing hydriodic acid is the reaction of iodine with hydrogen sulfide:H₂S + I₂ → 2 HI + S (b) How many grams of HI are produced from the reaction of 95.4 g of H₂S with excess I₂?