Limestone (CaCO 3 ) reacts with hydrochloric acid according to the equation CaCO 3 + 2 HCl ---> CaCl 2 + H 2 O + CO 2 . If 1.00 mol of CO 2 has a volume of 22.4 L under the reaction conditions, how many liters of gas can be formed by reaction of 2.35 g of CaCO 3 with 2.35 g of HCl? Which reactant is limiting?

Limestone reacts with hydrochloric acid. According to the following equation, if one mole of carbon dioxide has a volume of 22.4 L under the reaction conditions, how many liters of gas can be formed by reaction of 2.35 g of limestone with 2.35 g of hydrochloric acid which reactant is limiting. So the first step is to identify this limiting reactant by converting the masses of ca of our limestone and our hydrochloric acid to their corresponding number of moles using the num their molar mass and their corresponding mole ratios. For example, we have 2.35 g of our limestone are we first want to multiply this by its molar mass. And so if we look at the periodic table and we look at the molar masses of calcium, carbon and oxygen, we take that molar mass of that oxygen multiply that by three and then add them together, we get that one mole of this limestone is equal to 100.1 g. We then multiply it by their multiple ratio. And that means we need to multiply by the multiple ratio such that we look at the moles of our limestone, which is one mole since that is in our equation. And then we look for one of our products and we see that we have carbon dioxide that also has one mole ns. And we multiply this. And when we do so after crossing on our units, we arrive at an answer of 0.023 moles of our carbon dioxide. And we do the same thing for hydrochloric acid. And we also have 3.5 g of our hydrochloric acid, we multiply by its molar mass, such that one mole of a hydrochloric acid, which is the molar mass of chlorine. And the molar mass of hydrogen added up together is 36.46 g that we multiply by the multiple ratio. We see that in this equation, we have two moles of hydrochloric acid this time. So we have two moles of hydrochloric acid and the denominator is equal to that one mole of carbon dioxide in the numerator. And after our units cancel out, we arrive at 0.032 moles of carbon dioxide, the limiting reactant gives the lesser amount. So we see that we got a smaller number with calcium carbonate or our limestone. So our answer is either A or B now using dimensional analysis, we can calculate the liters of gas of carbon dioxide. And so from step one, we determined that we had our 0.023 moles of carbon dioxide and to convert 2 L of carbon dioxide, we can use the ratio that one mole of our gas is equal to 22.4 L. As given to us in the question stem, when our units canceled out, we arrive at an answer of 0.526 L of carbon dioxide. And we conclude that can conclude that our answer will be answer choice. A and with that, we have solved the problem overall, I hope it's helped. And until next time.

Ch.3 - Mass Relationships in Chemical Reactions

McMurry8th EditionChemistryISBN: 9781292336145Not the one you use?Change textbook

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McMurry 8th Edition

Ch.3 - Mass Relationships in Chemical Reactions

Problem 78

Chapter 3, Problem 78

Limestone (CaCO₃) reacts with hydrochloric acid according to the equation CaCO₃ + 2 HCl ---> CaCl₂ + H₂O + CO₂. If 1.00 mol of CO₂ has a volume of 22.4 L under the reaction conditions, how many liters of gas can be formed by reaction of 2.35 g of CaCO₃ with 2.35 g of HCl? Which reactant is limiting?

Verified step by step guidance

Calculate the molar mass of CaCO_3 and HCl using the periodic table: CaCO_3 = 40.08 (Ca) + 12.01 (C) + 3*16.00 (O) and HCl = 1.01 (H) + 35.45 (Cl).

Convert the mass of CaCO_3 and HCl to moles using their respective molar masses: moles = mass / molar mass.

Determine the stoichiometry of the reaction from the balanced equation: 1 mole of CaCO_3 reacts with 2 moles of HCl to produce 1 mole of CO_2.

Identify the limiting reactant by comparing the mole ratio of the reactants to the stoichiometric ratio from the balanced equation.

Calculate the volume of CO_2 produced using the moles of the limiting reactant and the given volume of 1 mole of CO_2 (22.4 L/mol).

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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 us to determine the amount of product formed or reactant consumed by using mole ratios derived from the coefficients in the balanced equation. In this case, stoichiometry will help us find out how much CO2 can be produced from the given amounts of CaCO3 and HCl.

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. To identify the limiting reactant, we compare the mole ratios of the reactants used in the reaction to the stoichiometric coefficients in the balanced equation. The reactant that produces the lesser amount of product is the limiting reactant.

Gas Volume and Molar Volume

At standard temperature and pressure (STP), one mole of an ideal gas occupies 22.4 liters. This concept is crucial for converting moles of gas produced in a reaction to volume. In this problem, knowing that 1.00 mol of CO2 corresponds to 22.4 L allows us to calculate the volume of gas produced from the moles of CO2 generated by the reaction of the limiting reactant.

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How many grams of the dry-cleaning solvent 1,2-dichloroethane (also called ethylene chloride), C₂H₄Cl₂, can be prepared by reaction of 15.4 g of ethylene, C₂H₄, with 3.74 g of Cl₂?

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