Question
Question: \( {P_4}(s) + 6C{l_2}(g) \to 4PC{l_3}(l) \) A reaction mixture initially contains \( 45.13g \) \( {P...
P4(s)+6Cl2(g)→4PCl3(l) A reaction mixture initially contains 45.13g P4 and 132.0g Cl2 . Once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant is left?
Solution
To solve this problem, we will first observe the number of moles of each reactant reacting with each other. Then we will convert the given mass of reactants into their number of moles. Therefore, using the stoichiometric calculations, we will see which reactant is left after the completion of the reaction.
Complete Step By Step Answer:
We are given with a balanced chemical equation which is written as:
P4(s)+6Cl2(g)→4PCl3(l)
We can see that one mole of P4 reacts with six moles of Cl2 . Now, we will find out the number of moles of each reactant using their given masses and molecular masses.
Given mass of P4 is 45.13g .
Given mass of Cl2 is 132.0g .
Molecular mass of P4 is 123.89g .
Molecular mass of Cl2 is 71g .
Now, the formula to calculate number of moles (n) in terms of given mass (m) and molecular mass (M.M.) is given as:
n=M.M.m
No. of moles of P4 are:
nP4=123.8945.13
nP4=0.3642
No. of moles of Cl2 are:
nCl2=71132.0
nCl2=1.8591
Six of Cl2 reacts with one mole of P4 . Therefore, Chlorine is the limiting reactant.
Therefore, 1.8591 of Cl2 reacts with:
=61×1.8591
=0.30985 moles of P4
Therefore, number of moles of P4 left after complete reaction is:
=0.3642−0.30985
=0.05435
Hence, the mass of P4 left after complete reaction is:
=0.05435×123.89
=6.7334g
Therefore, the mass of P4 left after complete reaction is 6.7334g .
Note:
The product PCl3 is known as Phosphorus trichloride. It has many applications such as it is used in the making of phosphorus oxychloride and this happens when Phosphorus trichloride is oxidized with oxygen. It is also used to make many chlorinated products such as Phosphoryl chloride, Phosphorus pentachloride and many more.