What is the temperature of 1.2 moles of helium gas at 2.57 ATM if it occupies 15.5L of volume
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Given parameters:
Mass of KClO₃ = 125g
Unknown:
Total mass of the products = ?
When KClO₃ is heated, it thermally decomposes to KCl and O₂ according to the chemical equation below;
2KClO₃ → 2KCl + 3O₂
All chemical equations obeys the law of conservation of matter and with this regard, we know that the amount of reactants used is the same as that of the product.
The total mass of the products must give us 125g according to this law of conservation of matter.
Now to find the masses of each product,
- Find the number of moles of the given reactant:
Number of moles =
molar mass of KClO₃ = 39 + 35.5 + 3(16) = 122.5g/mol
So number of moles of KClO₃ = = 1.02moles
2. Now, using this number of moles, find the number of moles of the products using this value;
2 moles of KClO₃ produced 2 moles of KCl
1.02 moles of KClO₃ will also produce 1.02moles of KCl
2 moles of KClO₃ produced 3 moles of O₂
1.02 moles of KClO₃ will produce mole = 1.53 moles of O₂
3. Now find the masses of each product;
Mass = number of moles x molar mass
molar mass of KCl = 39 + 35.5 = 74.5g/mol
molar mass of O₂ = 16 x 2 = 32g/mol
Mass of KCl = 74.5 x 1.02 = 75.99g
Mass of O₂ = 32 x 1.53 = 48.96g
Total mass of products = mass of KCl + Mass of O₂ = 75.99g + 48.96g
= 124.95g
This value is approximately the same as that of mass of KClO₃