Question

The atmosphere contained approximately 115 ppmv of anthropogenic co2(g) in 2010. if 1 ppmv- co2(g) is equivalent to 2184.82 tg-c (teragrams, or 1012 g, of atomic carbon), calculate how many metric tonnes of quicklime [cao(s)] were needed to fix

Answer 1

Answer is: 3,2 · 10¹¹ t of quicklime.
Chemical reaction: CO₂ + CaO → CaCO₃.
m(CO₂) = 2184,82·10¹² · 115 = 2,51 · 10¹⁷ g.
n(CO₂) = m(CO₂) ÷ M(CO₂)
n(CO₂) = 2,51 · 10¹⁷ g ÷ 44 g/mol = 5,7 · 10¹⁵ mol.
from reaction n(CO₂) : n(CaO) = 1 : 1.
n(CaO) = 5,7· 10¹⁵ mol.
m(CaO) = 5,7· 10¹⁵ mol · 56 g/mol = 3,2 · 10¹⁷ g = 3,2 · 10¹¹ t.
n - amount of substance.

Answer 2

Answer is: 3,2 · 10¹¹ t of quicklime.Â
Chemical reaction: COâ‚‚ + CaO → CaCOâ‚.
m(CO₂) = 2184,82·10¹² · 115 = 2,51 · 10¹ⷠg.
n(COâ‚‚) = m(COâ‚‚)Â Ă· M(COâ‚‚)
n(CO₂) = 2,51 · 10¹ⷠg ÷ 44 g/mol = 5,7 · 10¹ⵠmol.
from reaction n(COâ‚‚) : n(CaO) = 1 : 1.
n(CaO) = 5,7· 10¹ⵠmol.
m(CaO) = 5,7· 10¹ⵠmol · 56 g/mol = 3,2 · 10¹ⷠg = 3,2 ·
10¹¹ t.
n - amount of substance