Suppose 3.95% of 521 g PCl₅ (208.2 g/mol)initially present in a 10.0 L reaction vessel at 300 $\circ$ C decomposes to form PCl₃ (137.3 g/mol)and Cl₂ (70.90 g/mol).What is the resulting pressure of PCl₃? PCl₅(g)$\to$ PCl₃(g)+ Cl₂(g)

Question

Quizplus · Accepted Answer

To find the pressure of PCl₃, first calculate the moles of PCl₅ decomposed: 3.95% of 521 g is 20.56 g. Convert this to moles: 20.56 g / 208.2 g/mol = 0.0988 mol. Since the decomposition is 1:1, moles of PCl₃ formed = 0.0988 mol. Use PV=nRT to find pressure: P = (nRT)/V. With R = 0.0821 L·atm/mol·K and T = 300 K, P = (0.0988 mol * 0.0821 * 300 K) / 10.0 L = 0.243 atm.

Suppose 395% of 521 G PCl5 (208 ∘\circ∘ C Decomposes to Form PCl3 (137

Suppose 395% of 521 G PCl₅ (208 $\circ$ C Decomposes to Form PCl₃ (137