Answer to Question #73722 in General Chemistry for aleia

Question

Consider the balanced chemical reaction shown below.
4 PH3(g) + 8 O2(g) 6 H2O(l) + 1 P4O10(s)

In a certain experiment, 9.141 g of PH3(g) reacts with 5.649 g of O2(g).
(a) Which is the limiting reactant? (Example: type PH3 for PH3(g))

is the limiting reactant.


(b) How many grams of H2O(l) form?

g of H2O(l) form.


(c) How many grams of P4O10(s) form?

g of P4O10(s) form.


(d) How many grams of the excess reactant remains after the limiting reactant is completely consumed?

g of excess reactant remain.

Answer 1

4PH3(g) + 8O2(g)=6H2O(l) + P4O10(s)
.m(PH3)=9.141 g
.m(O2) =5.649 g
Limiting reactant=?
.m(H2O)=?
.m(P4O10)=?
.m(the excess reactant)=?
We find the amount of substance
.n(PH3)= m(PH3)/Mr(PH3)=9.141/34=0.27 mol
Mr(PH3)= Ar(P)+3 Ar(H)=31+3=34
Mr (O2)=2Ar(O)=2*16=32
.n(O2)=m(O2)/Mr(O2)=5.649/32=0.18 mol
a)O2 is a limiting reactant
PH3 is in excess
b).n(H2O)=6/8 n(O2)=6/8 *0.18=0.135 mol
.m(O2)=n(O2)* Mr(O2)=0.135 *32=4.32 g
c).n(P4O10)= 1/8 n(O2)=1/8 *0.18=0.0225 mol
.m(P4O10)=n*Mr(P4O10)=0.0225 *284=6.39 g
Mr(P4O10)=4Ar(P)+10Ar(O)=4 *31+16*10=284
d) Find the mass of the excess reactant)
.n(PH3) for reaction=1/2 n(O2)=1/2 *0.18=0.09 mol
.n(PH3) as excess reactant= n (PH3)geneal-n(PH3)for reaction=0.27-0.09=0.18 mol
.m (PH3) =n Mr(PH3)=0.18*34=6.12 g