Question

Given the following balanced equation Ca₃P2 + 6 H₂O ----> 3 Ca(OH) + 2 PH_3. a) What is the theoretical yield (in grams) of PH₃ produced from 125g each of Ca₃P₂ and H₂O? b) What is the limiting reagent? c) What is the percent yield if only 38.9g of PH₃ are produced?

Answer 1

a)

Molar mass of Ca3P2 = 3*MM(Ca) + 2*MM(P)

= 3*40.08 + 2*30.97

= 182.18 g/mol

mass of Ca3P2 = 125.0 g

we have below equation to be used:

number of mol of Ca3P2,

n = mass of Ca3P2/molar mass of Ca3P2

=(125.0 g)/(182.18 g/mol)

= 0.6861 mol

Molar mass of H2O = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass of H2O = 125.0 g

we have below equation to be used:

number of mol of H2O,

n = mass of H2O/molar mass of H2O

=(125.0 g)/(18.016 g/mol)

= 6.938 mol

we have the Balanced chemical equation as:

Ca3P2 + 6 H2O ---> 2 PH3 + 3 Ca(OH)2

1 mol of Ca3P2 reacts with 6 mol of H2O

for 0.6861 mol of Ca3P2, 4.1168 mol of H2O is required

But we have 6.9383 mol of H2O

so, Ca3P2 is limiting reagent

we will use Ca3P2 in further calculation

Molar mass of PH3 = 1*MM(P) + 3*MM(H)

= 1*30.97 + 3*1.008

= 33.994 g/mol

From balanced chemical reaction, we see that

when 1 mol of Ca3P2 reacts, 2 mol of PH3 is formed

mol of PH3 formed = (2/1)* moles of Ca3P2

= (2/1)*0.6861

= 1.372 mol

we have below equation to be used:

mass of PH3 = number of mol * molar mass

= 1.372*33.99

= 46.6 g

Answer: 46.6 g

b)

Ca3P2 is limiting reagent

c)

% yield = actual mass*100/theoretical mass

= 38.9*100/46.6

= 83.4 %

Answer: 83.4 %