To determine the volume of oxygen gas produced when 0.702 g of KCIO3 decomposes at STP, we first need to write the balanced chemical equation for the decomposition of KCIO3:
2KClO3 -> 2KCl + 3O2
From the balanced equation, we see that 2 moles of KCIO3 produce 3 moles of O2. Now we can calculate the molar mass of KCIO3:
K: 39.10 g/mol
Cl: 35.45 g/mol
O: 16.00 g/mol
Molar mass of KCIO3 = (39.10 + 35.45 + 3(16.00)) g/mol
Molar mass of KCIO3 = 122.55 g/mol
Next, we can calculate the number of moles of KCIO3 in 0.702 g:
Number of moles = 0.702 g / 122.55 g/mol
Number of moles = 0.00573 mol
Since 2 moles of KCIO3 produce 3 moles of O2, we can determine the number of moles of O2 produced:
Number of moles of O2 = 0.00573 mol KCIO3 * (3 mol O2 / 2 mol KCIO3)
Number of moles of O2 = 0.00860 mol O2
Now, we can use the ideal gas law to calculate the volume of O2 at STP (Standard Temperature and Pressure, which is 0°C and 1 atm):
PV = nRT
V = nRT / P
V = (0.00860 mol)(0.08206 L.atm/mol.K)(273 K) / 1 atm
V = 0.189 L
Therefore, the volume of oxygen gas produced when 0.702 g of KCIO3 decomposes at STP is 0.189 liters.
what volume pf oxygen gas is produced when 0.702 g of KCIO3 decomposes at STP
1 answer