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Ch.22 - The Main Group Elements
Chapter 22, Problem 70

The industrial production of hydriodic acid takes place by treatment of iodine with hydrazine (N2H4): 2 I2 + N2H4 --> 4 HI + N2 (a) How many grams of I2 are needed to react with 36.7 g of (N2H4)?

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1
Determine the molar mass of N2H4 by adding the atomic masses of nitrogen and hydrogen.
Calculate the number of moles of N2H4 using the given mass (36.7 g) and its molar mass.
Use the stoichiometry of the balanced chemical equation to find the moles of I2 needed. According to the equation, 2 moles of I2 react with 1 mole of N2H4.
Determine the molar mass of I2 by adding the atomic masses of iodine.
Calculate the mass of I2 required by multiplying the moles of I2 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 branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It allows us to calculate the amounts of substances consumed and produced in a reaction based on balanced chemical equations. Understanding stoichiometry is essential for determining how much of one substance is needed to react with a given amount of another.
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Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all the atoms in a molecule. Knowing the molar mass of reactants and products is crucial for converting between grams and moles, which is necessary for stoichiometric calculations in chemical reactions.
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Balanced Chemical Equation

A balanced chemical equation represents a chemical reaction with equal numbers of each type of atom on both sides of the equation. It ensures the law of conservation of mass is upheld, meaning that matter is neither created nor destroyed in a reaction. In this question, the balanced equation provides the mole ratios needed to relate the amounts of iodine (I2) and hydrazine (N2H4) involved in the reaction.
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