Chlorine is prepare by the reaction 2KmnO4 + 16Hcl 2Kcl + 2mncl2 + 5cl2 + 8H2O what mass of KmnO4 is required to prepare 500cm3 of chlorine gas at 73mmHg pressure at 25oC?
To calculate the mass of KMnO4 required to prepare 500 cm3 of chlorine gas at 73 mmHg pressure and 25oC, we need to use the ideal gas law equation.
The ideal gas law equation is given by:
PV = nRT,
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin.
First, we need to convert the given volume from cm3 to L:
V = 500 cm3 = 500/1000 L = 0.5 L.
Next, we need to convert the given pressure from mmHg to atm:
P = 73 mmHg = 73/760 atm.
The temperature is already given in oC, so we need to convert it to Kelvin by adding 273.15:
T = 25 oC + 273.15 = 298.15 K.
Now, we can rewrite the ideal gas law equation as:
(P * V)/(R * T) = n.
Since we want to find the number of moles of chlorine gas (n), we rearrange the equation:
n = (P * V)/(R * T).
Substituting the given values, we have:
n = (0.5 atm * 0.5 L) / (0.0821 L·atm/(mol·K) * 298.15 K).
Simplifying this equation, we find:
n ≈ 0.00945 mol.
From the balanced chemical equation: 2 KMnO4 + 16 HCl → 2 KCl + 2 MnCl2 + 5 Cl2 + 8 H2O,
we can see that 5 moles of Cl2 are produced for every 2 moles of KMnO4.
Therefore, the number of moles of KMnO4 required to produce 0.00945 mol of Cl2 is:
moles of KMnO4 = (0.00945 mol Cl2 * 2 mol KMnO4) / 5 mol Cl2.
Calculating this, we find:
moles of KMnO4 ≈ 0.00378 mol.
Finally, to find the mass of KMnO4 required, we need to multiply the number of moles by its molar mass, which can be obtained from the periodic table. The molar mass of KMnO4 is approximately 158.04 g/mol.
mass of KMnO4 = moles of KMnO4 * molar mass of KMnO4.
Substituting the values, we get:
mass of KMnO4 ≈ 0.00378 mol * 158.04 g/mol ≈ 0.597 g.
Therefore, approximately 0.597 grams of KMnO4 is required to prepare 500 cm3 of chlorine gas at 73 mmHg pressure and 25oC.