a)

CO_2+Ca(OH)_2->CaCO_3+H_2O

->V_{\text{khí thoát}}=V_{CO}=2,24(l)

->n_{CO}={2,24}/{22,4}=0,1(mol)

->\%m_{CO}={0,1.28}/{7,2}.100\%=38,89\%

->\%m_{CO_2}=100-38,89=61,11\%

b)

n_{CO_2}={7,2-0,1.28}/{44}=0,1(mol)

->M_C={7,2}/{0,1+0,1}=36(g//mol)

->d_{C//kk}={36}/{29}=1,24

c)

n_{CuO}={8,8}/{80}=0,11(mol)

CuO+CO  $$\xrightarrow{t^o}$$  Cu+CO_2

Do 0,11>0,1->H tính theo CO.

n_{CO\ pu}=0,1.80\%=0,08(mol)

Theo PT: n_{CuO\ pu}=n_{Cu}=n_{CO\ pu}=0,08(mol)

->m_{CR}=(0,11-0,08).80+0,08.64=7,52(g)