Question

For the following reaction, 15.2 grams of phosphorus (P₄) are allowed to react with 40.0 grams of chlorine gas .

phosphorus (P₄)(s) + chlorine(g) phosphorus trichloride(l)

What is the maximum mass of phosphorus trichloride that can be formed?  grams

What is the FORMULA for the limiting reagent?

What mass of the excess reagent remains after the reaction is complete?  grams

Answer 1

2/ - Reaction involved i n the 2 Pu (s) + 6 Cla (gs - upala (1) Moles of Dhoefhorus - Mass in gm. 15.2 -0.1226 Molar Mass 724 Moles of chlorine - yo 0.563. : I mole of Pu requiree 6 moles of cha - 01226 6 x 0.1226= 0.7356 motel As for fully Consumption of we need 0.7356 motes of cl bect we have only 0.563 moles of cl . So, Py won't fully Consumed. ich would be limiting reagent, ... Moles of Pcl3 formed - 4x Moles of ch o = 4, x 0.563 - 0.375. Mass of Pcs formed = 0.375x Molare mass of Pcl3 =0.375x 137.33 = 51.5gm & Moles of Py Censimed - Ix Moles of C2 -)x0563 +0.094 Moles of Py left = 0.1226-0.ogy = 0.0287. Mass 4 Py left = 0.0287 x 124= 3.567 gm :