$5/a)m_{CO_2}=\dfrac{4,5.10^{23}}{6.10^{23}}.44=33g$
$b)m_{CO_2}=1,5.44=66g$
$CO_2$ có $1,5.(6.10^{23})=9.10^{23}$ hạt
$6/a)$ $CaCO_3$ có $0,75.(6.10^{23})=4,5.10^{23}$ hạt
$m_{CaCO_3}=0,75.100=75g$
$b)H_2SO_4$ có $1,5.(6.10^{23})=9.10^{23}$ hạt
$m_{H_2SO_4}=1,5.98=147g$