Given the reaction: p4 (l) + 6 cl2 (g) → 4 pcl3 (l) if the percent yield is 82%, what mass of p4 is required to obtain 2.30 g pcl3 (cl2 in excess) ?

The mass of P4 required to obtain 2.30 PCl3 0.694 g

calculation

p4 + 6 Cl2→ 4 pcl3

step 1:calculate the moles of PCl3

moles = mass/molar mass

from periodic table the molar mass of PCl3 = 31 + (3 x35.5) = 137.5 g/mol

moles = 2.30 / 137.5 = 0.0167 moles

step 2: use the mole ratio to find the moles of P4

from equation above the mole ratio of p4 : PCl3 is 1:3 therefore the moles of p4

= 0.0167 x1/3 = 0.0056 moles

step 3: find the mass of P4

mass = moles x molar mass

from periodic table the molar mass of P4 = 31 x4 = 124 g/mol

mass = 0.0056 x 124 = 0.694 g