when magnesium chlorate under goes decompostion, oxgen gas is realsed in the following reaction:
2KClO3→ 2KCl + 3O2
repeat 2KClO3→ 2KCl + 3O2
when 3.50 grams of potassuim chlorate is decomposed at STP, how many litters of oxgen will be produced?
chemistry - Da
Katie,
That (3/2) in moles O2 to moles KClO3
is the 2 KClO3 for 3 moles O2 in your balanced equation
there is a 2 a 2 not 1 but 2KClO3→ 2KCl + 3O2
To determine the amount of oxygen gas produced when 3.50 grams of potassium chlorate (KClO3) is decomposed at STP, we need to use stoichiometry.
1. First, we need to calculate the number of moles of potassium chlorate using its molar mass. The molar mass of KClO3 is calculated as follows:
Molar mass (KClO3) = molar mass (K) + molar mass (Cl) + 3 * (molar mass (O))
= 39.10 g/mol + 35.45 g/mol + 3 * (16.00 g/mol)
= 122.55 g/mol
Now, we can calculate the number of moles of potassium chlorate:
Number of moles = Given mass (g) / Molar mass (g/mol)
= 3.50 g / 122.55 g/mol
≈ 0.0285 mol
2. Next, we use the balanced chemical equation to relate the number of moles of potassium chlorate to the number of moles of oxygen gas.
From the equation, we see that 2 moles of KClO3 produce 3 moles of O2.
Therefore, 0.0285 moles of KClO3 will produce (3/2) * 0.0285 = 0.0427 moles of O2.
3. Finally, we convert the number of moles of oxygen gas to liters at STP using the ideal gas law. At STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies approximately 22.4 liters.
Number of liters of oxygen gas = Number of moles * 22.4 L/mol
= 0.0427 mol * 22.4 L/mol
≈ 0.956 liters
Therefore, when 3.50 grams of potassium chlorate is decomposed at STP, approximately 0.956 liters of oxygen gas will be produced.
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