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, 3.981 g of Ca3P2(s) reacts with 4.412 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

MW Ca3P2 = 182.18

mol of Ca3P2 = mass/mol = 3.981 /182.18 = 0.02185 mol of Ca3P2

mol of H2O = mass/MW = 4.412/18 = 0.245111 mol ofH2O

ratio is 1:6 so

0.245111 mol of water will need 0.245111 /6 = 0.040851 mol of Ca3P2 which we do not have so

a)

Ca3P2 is limitin reactatn

b)

0.02185 mol of Ca3P2 react

since ratio is 1:2; that is 2 with respect to Ca(OH)2

then

0.02185 *3 = 0.06555 mol of Ca(OH)2 will be formed

mass = mol*MW = 0.06555*74 = 4.8507g of Ca(OH)2

c)

PH3

the ratio is 1:2 so

0.02185 mol of Ca3P2 will form 0.02185 *2 mol of PH3 = 0.0437mol of PH3

mass = mol*MW = 0.0437*33.99758 = 1.48569 g of PH3

d)

excess remains?

0.0437mol consumed so 0.040851*6 mol of H2O consumed

0.245106 mol of H2O consumed

H2o left = 0.245111 -0.0437 = 0.201411 mol of H2O left

mass = mol*MW =0.201411*18

mass = 3.625398 g of water left