CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

How many grams of PF5 can be formed from 9.46 g of PF3 and 9.42 g of XeF4 in the following reaction? (Molar mass of Xe =131 g/mol)
2PF3 + XeF42PF5 + Xe


Correct Answer
A
11.46 g
Your Answer
B
20.63 g
Your Answer
C
5.19 g
Your Answer
D
14.87 g

Solution

The correct option is A 11.46 g
2PF3+XeF42PF5+Xe
Moles of PF3=9.4688=0.1075 moles 
Moles of XeF4=9.42207=0.0455 moles
2 moles of PF3 react with 1 mole of XeF4
0.1075 moles will react with  12×0.1075=0.0538 moles of XeF4
Therefore, XeF4 is the limiting reagent.
1 mole of XeF4 produces 2 moles of PF5
0.0455 moles will produce 2×0.0455=0.091 moles of PF5
Mass of PF5=0.091×molar mass of PF5=(0.091×126)
Mass of PF5 = 11.46 g 

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
View More


People also searched for
View More



footer-image