16. From 200gm of CO2 ,10^21 molecules are removed. How many moles of CO2 are left?
Answers
ANHANITA BHOOSHAN HERE SEE BALVEER RETURNS MON TO FRI ON SONY SAB 7PM❤❤❤❤❤❤❤
Answer:
The method which we will use here consists of 2 steps :
Determine the weight of removed CO2
Subtract the weight of removed CO2 from initially available weight of CO2
Apply this method
Step 1.
No. of removed CO2 molecules = 10^21 ………(1)
Avogadro’s number = 6.022 × 10^23 ……….(2)
Moles of removed CO2 = No. of removed CO2 molecules ÷ Avogadro’s no.
Put the values of equations 1 and 2 in the above formula , we get ,
Moles of removed CO2 = 10^21 ÷ (6.022 × 10^23) = 0.166 × 10^-2 …………..(3)
Molecular weight of CO2 = 12 + (2×16) = 44g ………….…(4)
Weight of removed CO2 (g)= Moles of removed CO2 × molecular weight of CO2
Put the values of equations 3 and 4 in the above formula , we get ,
Weight of removed CO2 (g) = 0.166 × 10^-2 × 44 = 7.304 × 10^-2 g ………..…(5)
Step 2.
Initial weight of CO2 = 200 mg = 200 × 10^-3 g or 20 × 10^-2 g ……..………(6)
Left weight of CO2 = Initial CO2 weight - Removed CO2 weight
Put the values of equations 5 and 6 above , we get ,
Left CO2 weight = {(20 × 10 ^-2) - (7.304 × 10^-2)}g = 12.695 × 10^-2 g or 0.12695g
Explanation:
I HOPE YOU GET SOME HELP THIS ANS.