wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

5A2+2B42AB2+4A2B
The minimum mass of mixture of A2 and B4 in grams required to produce at least 1 kg of each product is:
(Given atomic mass of A=10; Atomic mass of B=120)

Open in App
Solution

5A2+2B42AB2+4A2B
Molar mass of A2=20 g/mol
molar mass of B4=480 g/mol
molar mass of AB2=250 g/mol
molar mass of A2B=140 g/mol
mass of AB2=1000 g
mass of A2B=1000 g
moles of AB2=1000250=4 mol
moles of A2B=1000140=7.14 mol

moles of AB2stoichiometric coefficient=42=2
moles of A2Bstoichiometric coefficient=7.144=1.785

In the case of the product, limiting product is the one that has the maximum ratio.

Here limiting product is AB2
Moles required of A2=5×42=10
So, mass of A2 needed =10×20=200 gm
moles of B4 needed =2×42=4
mass of B4 required = 4×480=1920 Total mass of mixture =2120 g

flag
Suggest Corrections
thumbs-up
3
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Stoichiometric Calculations
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon