In one reaction with a 90% yield, 200 mL of 2 M Na2CO3(aq),27gNO and a large excess O2 are allowed to react. 2Na2CO3(aq)+4NO(g)+O2(g)→4NaNO2(aq)+2CO2(g) what mass of NaNO2 is obtained based on experimental yield?
A
49.68 g
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
46.58 g
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
23.29 g
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
None of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A 49.68 g Number of moles of Na2CO3 =volume (L)×molarity=200×21000=0.4mol
Number of moles of NO =2730=0.9mol
Finding the limiting reagent: moles of Na2CO32=0.42=0.2 moles of NO4=0.94=0.225
Hence, Na2CO3 is the limiting reagent.
2 mol Na2CO3 gives 4 moles NaNO2
0.4 mol Na2CO3 gives =2×0.4=0.8molNaNO2 =0.8×69gNaNO2 =55.2gNaNO2 with 100% yield =55.2×0.90gNaNO2 with 90% yield =49.68 g