1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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 molLet us first solve this problem by writing the complete balanced reaction. 3BaCl2+2Na3PO4→Ba3(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.

Suggest Corrections
0
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program