. Calculate the mass of O2 produced if 3.450 g potassium chlorate is completely decomposed by heating in presence of a catalyst
1 answer:
<span>2 KClO3(s) → 3 O2(g) + 2 KCl(s)
</span><span>Note: MnO2 (Manganese Dioxide) is not part of the reaction. A catalyst lowers the activation energy and increases both forward and reverse reactions at equal rates.
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molar mass of KClO3 = 122.5
Moles of KClO3 = 3.45 / 122.55 = 0.028
Moles of O2 produce =
= 0.042 moles
molar mass of O2 = 32
so, mass of O2 = 32 x 0.042 = 1.35 g
</span><span>Note: MnO2 (Manganese Dioxide) is not part of the reaction. A catalyst lowers the activation energy and increases both forward and reverse reactions at equal rates.
</span>
molar mass of KClO3 = 122.5
Moles of KClO3 = 3.45 / 122.55 = 0.028
Moles of O2 produce =
= 0.042 moles
molar mass of O2 = 32
so, mass of O2 = 32 x 0.042 = 1.35 g
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0.6137 g of KHP contains 1.086 × 10^21 acidic protons.
Number of moles of KHP = mass of KHP/molar mass of KHP
Molar mass of KHP = 204.22 g/mol
Mass of KHP = 0.6137 g
Number of moles of KHP = 0.6137 g/204.22 g/mol = 0.003 moles of KHP
Now, 1 each molecule of KHP contains 1 acidic proton.
For 0.003 moles of KHP there are; 0.003 × 1 × NA
Where NA is Avogadro's number.
So; 0.003 moles of KHP contains 0.003 × 1 × 6.02 × 10^23
= 1.086 × 10^21 acidic protons.
Learn more: brainly.com/question/16672114
