Oxygen can be produced in the laboratory by the reaction: 2KClO₃(s) $\rightarrow$ 2KCl(s) + 3O₂(g)
How many moles of potassium chlorate are needed to produce 257 mL of oxygen,collected over water at 14^$\circ$C and 97.6 kPa? The vapor pressure of water at 14^$\circ$C is
1.60 kPa.

Question

Oxygen can be produced in the laboratory by the reaction: 2KClO₃(s)

\rightarrow

2KCl(s) + 3O₂(g)
How many moles of potassium chlorate are needed to produce 257 mL of oxygen,collected over water at 14^$\circ$C and 97.6 kPa? The vapor pressure of water at 14^$\circ$C is
1.60 kPa.

Quizplus · Accepted Answer

First, calculate the pressure of dry oxygen: 97.6 kPa - 1.60 kPa = 96.0 kPa. Convert 257 mL to liters: 257 mL = 0.257 L. Use the ideal gas law: PV = nRT. Solve for n (moles of O2): n = PV/RT = (96.0 kPa)(0.257 L) / (8.314 L·kPa/mol·K)(287 K) = 0.01034 mol O2. From the reaction, 2 moles of KClO3 produce 3 moles of O2, so moles of KClO3 = (2/3) * 0.01034 mol O2 = 0.00689 mol KClO3.

Oxygen Can Be Produced in the Laboratory by the Reaction →\rightarrow→

Oxygen Can Be Produced in the Laboratory by the Reaction $\rightarrow$