Answer to Question #95968 in General Chemistry for Aleah

Question

Consider the reaction:

P4(s) + 6Cl2(g) → 4PCl3(l)

Given an initial mass of 13.73 g P4, an excess of Cl2, and assuming that all of the reactant is converted to product(s), and none is lost, calculate the mass (g) of PCl3 produced by the reaction.

Answer 1

According to the equation 1 mole of P₄ react with 6 mole of Cl₂ to produce 4 mole of PCl₃

excess amount of Cl₂ and 13.73 g P₄ present at initial state

Thus P₄ is the limiting reagent

13.73 g P₄ contain 0.11 mole of P₄ will produce 0.44 mole of PCl₃

which is 60.42 gm