CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

If 0.5 moles of BaCl2 is mixed with 0.1 moles of Na3PO4, the maximum amount of Ba3(PO4)2 that can be formed is

A
0.7 mol
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.5 mol
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
0.2 mol
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0.05 mol
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 0.05 mol
Let us first solve this problem by writing the complete balanced reaction.
3BaCl2+2Na3PO4Ba3(PO4)2+6NaCl

We can see that the moles of BaCl2 used are 32 times the moles of Na3PO4.

Therefore, to react with 0.1 mol of Na3PO4, the moles of BaCl2 required would be 0.1×32=0.15.

Since the amount of BaCl2 is 0.5 mol, we can conclude that Na3PO4, is the limiting reagent.

Therefore, moles of Ba3(PO4)2 formed are 0.1×12=0.05 mol.
Hence, (d) is the correct answer.

flag
Suggest Corrections
thumbs-up
0
BNAT
mid-banner-image