Answer
$9.5\times10^{2}\,g\,CO_{2}$
Work Step by Step
Heat produced by the barbeque =
$1.6\times10^{3}\,kJ\times\frac{100}{10}= 1.6\times10^{4}\,kJ$
Mass of $CO_{2}$ emitted=
$-1.6\times10^{4}\,kJ\times\frac{3\,mol\,CO_{2}}{-2217\,kJ}\times\frac{44.01\,g\,CO_{2}}{1\,mol\,CO_{2}}$
$=9.5\times10^{2}\,g$
