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Chapter 1: Problem 10

Fill in the blanks. The volume of \(\mathrm{CO}_{2}\) liberated at \(\mathrm{STP}\) is on thermal decomposition of \(84 \mathrm{~g}\) of sodium bicarbonate.

Short Answer

Expert verified
Question: Calculate the volume of CO₂ liberated at STP on thermal decomposition of 84 g of sodium bicarbonate. Answer: The volume of CO₂ liberated at STP on thermal decomposition of 84 g of sodium bicarbonate is 11.2 Liters.

Step by step solution

01

Calculate the number of moles of sodium bicarbonate

First, we need to determine the molar mass of sodium bicarbonate (NaHCO₃). Using the periodic table, we find: Molar mass of Na = 22.99 g/mol Molar mass of H = 1.01 g/mol Molar mass of C = 12.01 g/mol Molar mass of O = 16.00 g/mol Molar mass of NaHCO₃ = 22.99 + 1.01 + 12.01 + (3×16.00) = 84.01 g/mol Next, we will use the given mass of sodium bicarbonate (84 g) to find the number of moles: Number of moles = mass/molar mass = 84 g / 84.01 g/mol = 1 mole
02

Determine the balanced equation for thermal decomposition of sodium bicarbonate

The balanced chemical equation for the thermal decomposition of sodium bicarbonate is: 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂ Using stoichiometry, we can see that for every 2 moles of NaHCO₃, 1 mole of CO₂ is produced.
03

Calculate the number of moles of CO₂ produced

Using the stoichiometry from the balanced equation, we can determine the number of moles of CO₂ produced: 1 mole of NaHCO₃ × (1 mole of CO₂ / 2 moles of NaHCO₃) = 0.5 mole of CO₂
04

Convert moles of CO₂ to volume at STP

At STP (Standard Temperature and Pressure) conditions, 1 mole of any gas occupies 22.4 Liters. To find the volume occupied by 0.5 mole of CO₂ at STP, we can use the formula: Volume = Number of moles × molar volume at STP = 0.5 mole × 22.4 L/mol = 11.2 L So, the volume of CO₂ liberated at STP on thermal decomposition of 84 g of sodium bicarbonate is 11.2 Liters.

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