Physical Properties: Sodium bicarbonate is an odorless, white crystalline solid or fine powder. It has a slightly alkaline taste. Its density is 2.20 g mL-1 and it decomposes in temperatures above 50 ºC. The decomposition yields to sodium carbonate. It is highly soluble in water and poorly soluble in acetone and methanol. It is insoluble in ethanol.
Chemical Properties: Sodium bicarbonate is an amphoteric compounds, it means the compound has a character acids an basic at the same time. It is highly soluble in water, resulting in a slighty alkaline solution.
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Step 1: Write Imbalance Equation
CH₃CHO + O₂ → CO₂ + H₂O
Step 2: Balance Carbon Atoms:
There are 2 carbon atoms at reactant side and one at product side. So multiply CO₂ with 2 to balance them. i.e.
CH₃CHO + O₂ → 2 CO₂ + H₂O
Step 3: Balance Hydrogen Atoms:
There are 4 hydrogen atoms at reactant side and 2 Hydrogen atoms at product side. So, multiply H₂O by 2 to balance Hydrogen on both sides. i.e.
CH₃CHO + O₂ → 2 CO₂ + 2 H₂O
Step 4: Balance Oxygen Atoms:
There are 3 Oxygen atoms at reactant side and 6 Oxygen atoms at product side. In order to balance them multiply O₂ on reactant side by 2.5 (5/2). i.e
CH₃CHO + 5/2 O₂ → 2 CO₂ + 2 H₂O
Step 6: Eliminate Fraction:
Multiply overall equation by 2 to eliminate fraction. i.e.
2 CH₃CHO + 5 O₂ → 4 CO₂ + 4 H₂O
Given parameters:
Weight of hydrated sample = 20g
Temperature = 250°C
Weight after cooling = 16.5g
Unknown:
Weight of water lost from the sample = ?
Solution:
The weight of water lost from the sample;
Weight of water lost = Weight of hydrated sample - Weight of dry sample
Weight of water lost = 20g - 16.5g = 3.5g
% of water in the sample = 
Input parameters solve;
= 
=17.5%
Element Symbol Mass Percent
Sodium Na 27.367%
The heat of the reaction is an extensive property: it is proportional to the quantity of the quantity that reacts.
The change in enthalpy is a measured of the heat evolved of absorbed.
When the heat is released, the change in enthalpy is negative.
The reaction of 2 moles of Na develops 368.4 kj of energy.
Calculate the number of moles of Na in 1.90 g to find the heat released when this quantity reacts.
Atomic mass of Na: 23 g/mol
#mol Na = 1.90 g / 23 g/mol = 0.0826 mol
Do the ratios: [368.4 kj/2mol ] * 0.0826 mol = 15.21 kj.
Then the answer is that 15.21 kj of heat is released (evolved)
