Given the balanced equation, how many milliliters of oxygen gas at STP are released from the decomposition of 5.05 g of calcium chlorate?
See you last problem.Same process.
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To determine the amount of oxygen gas released from the decomposition of calcium chlorate, we need to follow these steps:
Step 1: Write and balance the chemical equation.
The decomposition of calcium chlorate can be represented by the balanced equation:
2 Ca(ClO3)2 -> 2 CaCl2 + 3 O2
Step 2: Calculate the molar mass of calcium chlorate (Ca(ClO3)2).
Ca: 1 atom x 40.08 g/mol = 40.08 g/mol
Cl: 2 atoms x 35.45 g/mol = 70.90 g/mol
O: 6 atoms x 16.00 g/mol = 96.00 g/mol
Molar mass of Ca(ClO3)2 = 40.08 + 70.90 + 96.00 = 206.98 g/mol
Step 3: Convert the given mass of calcium chlorate (5.05 g) to moles.
moles = mass / molar mass
moles = 5.05 g / 206.98 g/mol ≈ 0.024 moles
Step 4: Use the stoichiometry of the balanced equation to convert moles of calcium chlorate to moles of oxygen gas.
According to the balanced equation,
2 moles of Ca(ClO3)2 produce 3 moles of O2
So, for 0.024 moles of Ca(ClO3)2:
0.024 moles Ca(ClO3)2 x (3 moles O2 / 2 moles Ca(ClO3)2) = 0.036 moles of O2
Step 5: Convert moles of oxygen gas to volume using the ideal gas law equation.
PV = nRT
At STP (Standard Temperature and Pressure), T = 273 K, P = 1 atm, and R (the ideal gas constant) = 0.0821 L·atm/mol·K.
V = nRT / P
V = 0.036 mol x 0.0821 L·atm/mol·K x 273 K / 1 atm
V ≈ 0.82 L
Step 6: Convert liters to milliliters.
1 L = 1000 mL
Therefore, the volume of oxygen gas released from the decomposition of 5.05 g of calcium chlorate at STP is approximately 820 mL.