How many milliliters of oxygen gas at STP are released from heating 5.15g of calcium chlorate?
Ca(ClO3)2(s)→CaCl2(s)+3O2(g)
To find the number of milliliters of oxygen gas released, we can follow these steps:
Step 1: Convert the given mass of calcium chlorate (Ca(ClO3)2) to moles.
To do this, we need to calculate the molar mass of calcium chlorate.
Calcium (Ca) has a molar mass of approximately 40.08 g/mol.
Chlorine (Cl) has a molar mass of approximately 35.45 g/mol.
Oxygen (O) has a molar mass of approximately 16.00 g/mol.
Since there are 2 chlorine atoms and 6 oxygen atoms in the formula of calcium chlorate, the molar mass of Ca(ClO3)2 will be:
Molar mass = 1(40.08 g/mol) + 2(35.45 g/mol) + 6(16.00 g/mol)
Molar mass = 40.08 g/mol + 70.90 g/mol + 96.00 g/mol
Molar mass = 206.98 g/mol
Next, we can use the molar mass to convert the given mass of calcium chlorate to moles:
Number of moles = Mass / Molar mass
Number of moles = 5.15 g / 206.98 g/mol
Number of moles ≈ 0.0249 moles (rounded to 4 decimal places)
Therefore, 5.15 grams of calcium chlorate is approximately equal to 0.0249 moles.
Step 2: Use the mole ratio from the balanced equation to find the moles of oxygen gas.
From the balanced equation, we can see that 1 mole of calcium chlorate (Ca(ClO3)2) produces 3 moles of oxygen gas (O2).
So, the number of moles of oxygen gas produced will be:
Moles of oxygen gas = Moles of calcium chlorate × (3 moles O2 / 1 mole Ca(ClO3)2)
Moles of oxygen gas = 0.0249 moles × (3 moles O2 / 1 mole Ca(ClO3)2)
Moles of oxygen gas = 0.0748 moles (rounded to 4 decimal places)
Step 3: Use the ideal gas law to convert moles of oxygen gas to volume at STP (Standard Temperature and Pressure).
At STP, the temperature is 0°C or 273.15 K, and the pressure is 1 atm.
According to the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
Since we want to find the volume (V), rearrange the ideal gas law equation:
V = (nRT) / P
Substituting the values:
V = (0.0748 moles × 0.0821 L·atm/(mol·K) × 273.15 K) / 1 atm
V ≈ 1.84 L (rounded to 2 decimal places)
Therefore, approximately 1.84 milliliters of oxygen gas at STP are released from heating 5.15 grams of calcium chlorate.
To determine the number of milliliters of oxygen gas released from heating 5.15g of calcium chlorate (Ca(ClO3)2) at standard temperature and pressure (STP), we need to use the concept of moles and the molar volume of a gas at STP.
1. Start by calculating the number of moles of calcium chlorate (Ca(ClO3)2) using its molar mass, which can be found on the periodic table.
Molar mass of Ca(ClO3)2 = (1 * Atomic mass of Ca) + (2 * Atomic mass of Cl) + (6 * Atomic mass of O)
= (1 * 40.08 g/mol) + (2 * 35.45 g/mol) + (6 * 16.00 g/mol)
= 40.08 g/mol + 70.90 g/mol + 96.00 g/mol
= 206.98 g/mol
Moles of Ca(ClO3)2 = Mass of Ca(ClO3)2 / Molar mass of Ca(ClO3)2
= 5.15 g / 206.98 g/mol
≈ 0.0248 mol
2. Using the balanced chemical equation, we can determine the mole ratio between Ca(ClO3)2 and O2. According to the equation:
Ca(ClO3)2(s) → CaCl2(s) + 3O2(g)
The ratio is 1 mol of Ca(ClO3)2: 3 mol of O2.
3. Convert the moles of Ca(ClO3)2 to moles of O2 using the mole ratio:
Moles of O2 = Moles of Ca(ClO3)2 * (3 mol O2 / 1 mol Ca(ClO3)2)
= 0.0248 mol * (3 mol O2 / 1 mol Ca(ClO3)2)
= 0.0744 mol O2
4. Finally, we can use the molar volume of a gas at STP, which is approximately 22.4 L/mol, to convert the moles of O2 to milliliters:
Volume of O2 = Moles of O2 * 22.4 L/mol
= 0.0744 mol * 22.4 L/mol
≈ 1.6656 L
However, you requested the answer in milliliters, so we need to convert liters to milliliters:
Volume of O2 = 1.6656 L * (1000 mL / 1 L)
= 1665.6 mL
Therefore, approximately 1665.6 milliliters of oxygen gas would be released from heating 5.15g of calcium chlorate at STP.