Answer
(a) 0.350 g
(b) 0.221 g
(c) 8.15 g
Work Step by Step
(a) $178\times10^{-3}\,L\,CO_{2}\times\frac{1\,mol\,CO_{2}}{22.4\,L\,CO_{2}}\times\frac{44.01\,g\,CO_{2}}{1\,mol\,CO_{2}}$
$= 0.350\,g$
(b) $155\times10^{-3}\,L\,O_{2}\times\frac{1\,mol\,O_{2}}{22.4\,L\,CO_{2}}\times\frac{32.0\,g\,O_{2}}{1\,mol\,O_{2}}$
$=0.221\,g$
(c) $1.25\,L\,SF_{6}\times\frac{1\,mol\,SF_{6}}{22.4\,L\,SF_{6}}\times\frac{146.06\,g\,SF_{6}}{1\,mol\,SF_{6}}$
$=8.15\,g$
