Question In Description

Question

A scientist used the reaction below to generate KCLO3. How many grams of KCLO3 could be prepared from 8.75 L of Cl2 gas, if the gas was originally at a pressure of 1275 mmHg at a temperature of 57°C.

Show work, use gas stoicheometry

Answer 1

The given information allows us to use the ideal gas law to calculate the number of moles of Cl2 gas:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

We first need to convert the given pressure from mmHg to atm, and the temperature from Celsius to Kelvin:

P = 1275 mmHg * (1 atm / 760 mmHg) = 1.68 atm T = 57°C + 273.15 = 330.15 K

Now we can plug in the values and solve for n:

n = PV / RT = (1.68 atm * 8.75 L) / (0.0821 L·atm/mol·K * 330.15 K) ≈ 0.372 mol Cl2

Next, we can use the balanced chemical equation to determine the stoichiometric ratio between Cl2 and KCLO3:

2 KCl + 3 Cl2 + 3 H2O → 2 KClO3 + 6 HCl

For every 3 moles of Cl2, we can produce 2 moles of KCLO3. Therefore, the number of moles of KCLO3 produced is:

n(KCLO3) = (2/3) * n(Cl2) = (2/3) * 0.372 mol ≈ 0.248 mol

Finally, we can use the molar mass of KCLO3 to convert moles to grams:

m(KCLO3) = n(KCLO3) * M(KCLO3) = 0.248 mol * 122.55 g/mol ≈ 30.4 g

Thus, about 30.4 grams of KCLO3 could be prepared from 8.75 L of Cl2 gas at 1275 mmHg and 57°C.