What volume of 0.600 M HCl is required to react completely with 2.50 g of sodium hydrogen carbonate? \(\mathrm{NaHCO}_{3}(a q)+\mathrm{HCl}(a q) \rightarrow \mathrm{NaCl}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\)

Short Answer

Expert verified
The volume of 0.600 M HCl required is 0.0298 L, or 29.8 mL.

Step by step solution

01

Calculate the number of moles of NaHCO3

To calculate the moles of NaHCO3, use the formula: moles = mass (g) / molar mass (g/mol). The molar mass of NaHCO3 is approximately 84.01 g/mol. So, moles of NaHCO3 = 2.50 g / 84.01 g/mol.
02

Use the stoichiometry of the reaction

From the balanced equation, 1 mole of NaHCO3 reacts with 1 mole of HCl. Therefore, the moles of HCl needed equals the moles of NaHCO3.
03

Calculate the volume of HCl solution

Use the molarity formula: Molarity (M) = moles of solute / volume of solution (L). Rearrange to find the volume of the HCl solution: Volume (L) = moles of HCl / molarity of HCl. Substitute the values to calculate the volume required.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Molarity
Molarity is a measure of the concentration of a solute in a solution. It is defined as the number of moles of a solute divided by the volume of the solution in liters. The molarity concept is crucial when preparing solutions in chemistry and understanding reaction stoichiometry.

To express this mathematically, the formula for molarity (M) is:
\[ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \].

When solving problems that involve chemical reactions, molarity provides a bridge between the amount of substance and the volume of solution. For instance, to find the volume of a molarity solution required to react with a given mass of another substance, you can start by calculating the number of moles of the substance. Afterward, using the molarity of the other reactant, determine the volume of solution that contains an equivalent amount of moles required for complete reaction.
Navigating Chemical Reactions
Chemical reactions are the processes by which substances convert into new materials. They are represented by balanced chemical equations that show the relationship between reactants and products. The coefficients in a chemical equation provide the stoichiometry of the reaction, which is the proportion in which reactants combine and products form.

A concept intertwined with chemical reactions is the law of conservation of mass; it states that mass is neither created nor destroyed, leading to an equal number of atoms on both the reactant and product sides in a balanced equation. As a result, you can use stoichiometry to determine the amounts of reactants needed or products formed in a reaction.

In practice, this means that if a reaction states that 1 mole of substance A reacts with 1 mole of substance B, you will need equal moles of each for them to react completely. This ratio can be used to calculate unknown quantities if one of them is given, often through molarity calculations.
Molar Mass Calculation
The molar mass of a compound is a fundamental concept that connects the macroscopic world we can measure to the microscopic world of atoms and molecules. It is the mass of one mole of a given substance, typically expressed in grams per mole (g/mol).

Molar mass is calculated by summing the atomic masses of all the atoms present in the molecule, as found on the periodic table. For example, the molar mass of sodium hydrogen carbonate (NaHCO3) is determined by adding the atomic masses of one sodium (Na), one hydrogen (H), one carbon (C), and three oxygen (O) atoms.

Once you have the molar mass, calculating the number of moles of a substance is straightforward using the formula:
\[ \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \].

Knowing the number of moles allows you to analyze chemical reactions in terms of quantity, which is essential when precisely measuring reactants and products in lab settings or examining the yields of reactions.

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Most popular questions from this chapter

Toluene, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}\), is oxidized by air under carefully controlled conditions to benzoic acid, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\), which is used to prepare the food preservative sodium benzoate, \(C_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{Na} .\) What is the percent yield of a reaction that converts 1.000 kg of toluene to 1.21 kg of benzoic acid? \(2 \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}+3 \mathrm{O}_{2} \longrightarrow 2 \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}+2 \mathrm{H}_{2} \mathrm{O}\)

What volume of 0.0105-M HBr solution is required to titrate 125 mL of a 0.0100-M Ca(OH)_solution? \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)+2 \mathrm{HBr}(a q) \rightarrow \mathrm{CaBr}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

Outline the steps needed to determine the limiting reactant when 0.50 mol of \(\mathrm{Cr}\) and \(0.75 \mathrm{mol}\) of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) react according to the following chemical equation. \(2 \mathrm{Cr}+2 \mathrm{H}_{3} \mathrm{PO}_{4} \longrightarrow 2 \mathrm{CrPO}_{4}+3 \mathrm{H}_{2}\) Determine the limiting reactant.

Balance the following equations: (a) \(\mathrm{PCl}_{5}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{POCl}_{3}(l)+\mathrm{HCl}(a q)\) (b) \(\mathrm{Cu}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(I)+\mathrm{NO}(g)\) (c) \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(s) \longrightarrow \mathrm{HI}(s)\) (d) \(\operatorname{Fe}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) (e) \(\mathrm{Na}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g)\) (f) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(s) \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{3}(s)+\mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (g) \(\mathrm{P}_{4}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{3}(l)\) (h) \(\mathrm{PtCl}_{4}(s) \longrightarrow \operatorname{Pt}(s)+\mathrm{Cl}_{2}(g)\)

The toxic pigment called white lead, \(\mathrm{Pb}_{3}(\mathrm{OH})_{2}\left(\mathrm{CO}_{3}\right)_{2}\), has been replaced in white paints by rutile, TiO_ \(.\) How much rutile (g) can be prepared from 379 g of an ore that contains \(88.3 \%\) ilmenite (FeTiO \(_{3}\) ) by mass? \(2 \mathrm{FeTiO}_{3}+4 \mathrm{HCl}+\mathrm{Cl}_{2} \longrightarrow 2 \mathrm{FeCl}_{3}+2 \mathrm{TiO}_{2}+2 \mathrm{H}_{2} \mathrm{O}\)

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