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How many grams of phosphine (PH3) can form when 37.5 g of
Chapter 6, Problem 45P(choose chapter or problem)
How many grams of phosphine (PH3) can form when \(37.5 g\) of phosphorus and \(83.0 L\) of hydrogen gas react at STP?
\(P_{4}(s)+H_{2}(g) \rightarrow P H_{3}(g)\) [unbalanced]
Equation Transcription:
Text Transcription:
37.5 g
83.0 L
P_4(s) + H2(g) rightarrow PH_3(g)
Questions & Answers
QUESTION:
How many grams of phosphine (PH3) can form when \(37.5 g\) of phosphorus and \(83.0 L\) of hydrogen gas react at STP?
\(P_{4}(s)+H_{2}(g) \rightarrow P H_{3}(g)\) [unbalanced]
Equation Transcription:
Text Transcription:
37.5 g
83.0 L
P_4(s) + H2(g) rightarrow PH_3(g)
ANSWER:
Solution 45P
Here we have to find out mass of phosphine (PH3) can form in grams .
Step 1
Given:
Mass of phosphorus = 37.5 g
Volume of Hydrogen gas = 83.0 L
The balanced chemical equation for the reaction between phosphorous and hydrogen is given below,
P4(s) + 6H2(g) → 4PH3(g)
In order to find out the mass of PH3, 1st we have to find out the limiting reactant, then by using the limiting reactant we can calculate the mass of PH3.
PH3 from P4:-
In the above reaction it has been found that 4 mole of PH3 has formed from 1 mole of P4.
Thus the mole of PH3 formed can be calculated as,
Mass of P4
= 37.5 g of P4
= 1.21 mol PH3
PH3 from H2:-
From the above balanced equation it has been found that 4 mole of PH3 has formed from 6 mole of H2.
1st we will convert of hydrogen into mole as follows,
It is known that 1 mole of hydrogen = 24L
Thus 83.0 L of hydrogen will be equal to = 83.0 L
= 3.705 mole
Thus the mole of PH3 formed can be calculated as,
Mole of H2
= 3.705 mole of H2
= 2.47 mol of PH3
The mole of PH3 formed from P4 is less hence it is the limiting reagent.