A You have 1.249 g of a mixture of NaHCO3 and Na2 CO3 You find that 12.0 mL of 1.50 M HCl is req

step 1 This is a fairly lengthy and complicated problem. It ultimately is going to require two equation

A You have 1.249 g of a mixture of NaHCO3 and Na2 CO3 You...

A You have $1.249 \mathrm{g}$ of a mixture of $\mathrm{NaHCO}_{3}$ and $\mathrm{Na}_{2} \mathrm{CO}_{3}$ You find that $12.0 \mathrm{mL}$ of $1.50 \mathrm{M} \mathrm{HCl}$ is required to convert the sample completely to $\mathrm{NaCl}, \mathrm{H}_{2} \mathrm{O},$ and $\mathrm{CO}_{2}$ $\mathrm{NaHCO}_{3}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \longrightarrow$ $$ \mathrm{NaCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{CO}_{2}(\mathrm{g}) $$ $\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow$ $$ 2 \mathrm{NaCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{CO}_{2}(\mathrm{g}) $$ What volume of $\mathrm{CO}_{2}$ is evolved at $745 \mathrm{mm}$ Hg and $25^{\circ} \mathrm{C} ?$

A You have $1.249 \mathrm{g}$ of a mixture of $\mathrm{NaHCO}_{3}$ and $\mathrm{Na}_{2} \mathrm{CO}_{3}$ You find that $12.0 \mathrm{mL}$ of $1.50 \mathrm{M} \mathrm{HCl}$ is required to convert the sample completely to $\mathrm{NaCl}, \mathrm{H}_{2} \mathrm{O},$ and $\mathrm{CO}_{2}$ $\mathrm{NaHCO}_{3}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \longrightarrow$
$$
\mathrm{NaCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{CO}_{2}(\mathrm{g})
$$
$\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow$
$$
2 \mathrm{NaCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{CO}_{2}(\mathrm{g})
$$
What volume of $\mathrm{CO}_{2}$ is evolved at $745 \mathrm{mm}$ Hg and $25^{\circ} \mathrm{C} ?$

Key Concepts

Balanced Chemical Equations and Reaction Ratios

Balanced chemical equations provide the mole ratios needed to understand how different reactants interact, particularly when a compound reacts in more than one way (for example, using different stoichiometric ratios for different components of a mixture). Understanding these ratios is essential to relate the amounts of various substances in a reaction.

Ideal Gas Law

The Ideal Gas Law (PV = nRT) relates pressure, volume, temperature, and the number of moles of a gas. It is used to calculate the volume of gas produced or required under specific conditions, making it a fundamental tool when converting between moles of a gas and its corresponding volume at given temperature and pressure.

Stoichiometry

Stoichiometry is the calculation of the quantitative relationships between reactants and products in a chemical reaction. It involves using balanced chemical equations to determine the proportion of substances consumed and produced, which is critical when dealing with mixtures or multiple reactions.

Molarity and Volume Relationship

Molarity defines the concentration of a solution as moles of solute per liter of solution. By multiplying the molarity by the volume of the solution used, one can determine the number of moles of a reactant or reagent, which is a key step in quantitative chemical analysis.

Transcript

00:01 This is a fairly lengthy and complicated problem.

00:05 It ultimately is going to require two equations and two unknowns.

00:09 We also need to have some understanding of the ideal gas law.

00:14 So let's get started.

00:18 We're going to define x as the mass of sodium bicarbonate that we have and y as the mass of sodium carbonate that we have.

00:26 Then x plus y is going to be equal to the total mass that was given given to us of 1 .249 grams.

00:42 Then we know the volume of hcl at a particular molarity required to neutralize both the sodium carbonate and the sodium bicarbonate.

00:53 Multiplying the volume times the molarity will give us the moles of hcl that was required.

01:02 The moles of hcl is also going to be equal to the mass, that is sodium bicarbonate divided by its molar mass, the mass that is sodium bicarbonate we don't know, so that's x, and then we'll convert the moles of sodium bicarbonate into moles of hydrochloric acid.

01:23 This then is the moles of hydrochloric acid that is required to neutralize x moles of sodium bicarbonate.

01:32 That then is part of the total moles of hcl that we added.

01:36 Then to calculate the moles of hcl required to neutralize the sodium carbonate that's present, we have y grams of sodium carbonate divided by its molar mass to get mole sodium carbonate, and then looking at the stoichiometry of the reaction, we need 2 moles of hcl for every 1 mole of sodium carbonate.

01:57 Now we have two equations, this one and this one, and two unknowns, so we can substitute in for y, x, or 1 .249 minus x...

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