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\(2.0 \mathrm{~g}\) mixture of sodium carbonate and sodium bicarbonate on heating to constant weight gave \(224 \mathrm{~mL}\) of \(\mathrm{CO}_{2}\) at N.T.P. The \(\%\) weight of sodium bicarbonate in the mixture is: (a) 50 \(\square\) (b) 54 (c) 80 \(\square\) (d) 84

Short Answer

Expert verified
84% sodium bicarbonate in the mixture, option (d).

Step by step solution

01

Understanding the Problem

We have a mixture of sodium carbonate (\(\text{Na}_2\text{CO}_3\)) and sodium bicarbonate (\(\text{NaHCO}_3\)) which on heating gives carbon dioxide \((\text{CO}_2)\). Our task is to find the percentage by weight of sodium bicarbonate in this mixture. Given is the mass of the mixture as \(2.0\, \text{g}\) and the volume of \(\text{CO}_2\) produced is \(224\, \text{mL}\), at N.T.P.
02

Reaction Equations

Both \(\text{NaHCO}_3\) and \(\text{Na}_2\text{CO}_3\) produce \(\text{CO}_2\) on heating but we consider that \(\text{Na}_2\text{CO}_3\) does not decompose at temperatures typically used. The decomposition reaction for \(\text{NaHCO}_3\) is:\[ 2 \text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{CO}_2 + \text{H}_2\text{O} \]
03

Calculate Moles of \(\text{CO}_2\)

Since \(1\, \text{mol}\) of \(\text{CO}_2\) occupies \(22400\, \text{mL}\) at N.T.P., the moles of \(\text{CO}_2\) is:\[ \text{Moles of } \text{CO}_2 = \frac{224\, \text{mL}}{22400\, \text{mL/mol}} = 0.01\, \text{mol} \]
04

Relate Moles Back to \(\text{NaHCO}_3\)

From the reaction, \(2\) moles of \(\text{NaHCO}_3\) produce \(1\) mole of \(\text{CO}_2\). Therefore, \(0.01\, \text{mol}\) of \(\text{CO}_2\) will require:\[ 2 \times 0.01 = 0.02\, \text{mol} \, \text{of } \text{NaHCO}_3 \]
05

Calculate Mass of \(\text{NaHCO}_3\)

The molar mass of \(\text{NaHCO}_3\) is \(84\, \text{g/mol}\). The mass of \(\text{NaHCO}_3\) is:\[ 0.02\, \text{mol} \times 84\, \text{g/mol} = 1.68\, \text{g} \]
06

Calculate Percentage of \(\text{NaHCO}_3\)

The percentage by weight of \(\text{NaHCO}_3\) in the mixture is:\[ \frac{1.68\, \text{g}}{2.0\, \text{g}} \times 100\% = 84\% \]
07

Conclude the Solution

Therefore, the percentage by weight of sodium bicarbonate in the mixture is \(84\%\) which corresponds to option (d).

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Key Concepts

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

Sodium Carbonate
Sodium carbonate, also known as washing soda or soda ash, is a chemical compound with the formula \( \text{Na}_2\text{CO}_3 \). It is an important industrial product used in glass manufacturing, as a water softener in laundry detergents, and for pH regulation in various processes.

In the context of the problem, sodium carbonate is part of the original mixture but does not contribute to the carbon dioxide release at the temperatures used in typical lab conditions. This is because sodium carbonate does not decompose under these conditions, unlike sodium bicarbonate. Thus, its presence affects the weight of the mixture but not the volume of \( \text{CO}_2 \) produced.
  • Stable under typical heating conditions.
  • Does not decompose to release \( \text{CO}_2 \).
  • Adds to the mass of the mixture without affecting \( \text{CO}_2 \) volume.
Sodium Bicarbonate
Sodium bicarbonate, commonly known as baking soda, is a chemical compound with the formula \( \text{NaHCO}_3 \). It is widely used in baking, cleaning, and as an antacid.

In chemical reactions, sodium bicarbonate is known for its ability to decompose upon heating. When heated, it breaks down into sodium carbonate, carbon dioxide gas, and water vapor. This decomposition is represented by the equation: \[ 2 \text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{CO}_2 + \text{H}_2\text{O} \] In our exercise, the focus is on calculating how much of the original mixture consists of sodium bicarbonate. As sodium bicarbonate decomposes, it releases \( \text{CO}_2 \), which is measured. By determining the amount of \( \text{CO}_2 \) produced, we can trace back to find the moles and thus the mass of sodium bicarbonate present initially. Ultimately, this allows calculation of the percentage composition of the mixture.
  • Decomposes on heating, releasing \( \text{CO}_2 \) gas.
  • Essential for calculating gas volume in the context.
  • Key to determining the mass and percentage composition in the mixture.
Gas Volume at N.T.P.
N.T.P. stands for Normal Temperature and Pressure, which typically refers to conditions of \( 0^\circ \text{C} \) (273.15 K) and 1 atm pressure. Under these standardized conditions, one can observe consistent behavior in gas volumes, making calculations simpler and more predictable.

For example, at N.T.P., one mole of any ideal gas occupies a volume of 22.4 L (or 22400 mL). This knowledge is crucial in stoichiometric calculations when converting between moles of a gas and its volume.

In the problem at hand, the amount of carbon dioxide gas measured was 224 mL at N.T.P. By using the molar volume of a gas at these conditions, we were able to determine that this volume corresponds to 0.01 moles of \( \text{CO}_2 \). Understanding gas volume at N.T.P. ensures accuracy and uniformity in chemical calculations.
  • Provides reliable and uniform conditions for gas calculations.
  • Helps convert between moles and volume for gases.
  • Essential for calculating the moles of \( \text{CO}_2 \) from measured volume.

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Most popular questions from this chapter

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