Question

33.6g of an impure sample of sodium bicarbonate when heated strongly gave 4.4g of CO2. The % purity of NaHCO3would be:
a)25%
b)50%
c)75%
d)100%

Answer 1

see the attachment for your answer

Answer 2

Answer:

(b)50%

Explanation:

The balanced chemical reaction is:

2NaHCO3→→→→Na2CO3+H2O+CO2

Moles of CO2= 4.4/44=0.1 MOLES

∴Moles Of NaHCO3(That Reacted)=01×2= 0.2 Moles

∴Mass Of NaHCO3(That Reacted)=0.2*84=16.8 gm

Percent purity = (16.8÷33.6)×100 = 50%

Thus the answer is 50%

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