Question

Consider the balanced chemical reaction shown below. 1 Ca3P2(s) + 6 H2O(l) 3 Ca(OH)2(s) + 2 PH3(g) In a certain experiment, 9.055 g of Ca3P2(s) reacts with 2.224 g of H2O(l). (A)Which is the limiting reactant? (Example: type Ca3P2 for Ca3P2(s)) (B)How many grams of Ca(OH)2(s) form? (C)How many grams of PH3(g) form? (D)How many grams of the excess reactant remains after the limiting reactant is completely consumed?

Answer 1

The reactino is

Ca3P2(s) + 6 H2O(l) ----->3 Ca(OH)2(s) + 2 PH3(g) ( 1 mole of Ca3P2 reacts with 6 moles of water to give rise to 3 moles of Ca(OH)2 and 2 moles of PH3.

Atomic weights : P =31 and Ca=40 H= 1 and O=16

Molecular weights : Ca3P2= 40*3+2*31= 182   H2O= 2+16=18   Ca(OH)2= 40+2*(1+16)= 74 and PH3= 31+3=34

Moles = mass/ Molecular weight

Moles : Ca3(PO4)2= 9.055/182=0.049753    water= 2.224/18=0.124

as per stoichiometry, molar ratio of reactants = 1:6

as per the given msses, molar ratio = 0.049753: 0.124= 1:2.5

So the ratio suggests water to be the limiting reactant.

Limiting reactant influences the extent of formation of Ca(OH)2

6*18 gm =108 gms of water gives 3*74=222 gms of Ca(OH)2

2.224gms gives 2.224*222/108=4.57 gms of Ca(OH)2

also

108 gm of water requires 182 gm of Ca3P2

2.224 gm of water requires 2.224*182/108=3.75 gm of Ca3P2

supplied Ca3P2= 9.055 gm

Ca3P2 remaining after the reactions= 9.055-3.75=5.305 gms