\(m_{CaCO_3}=500.80\%=400\left(g\right)\)
`=>` \(n_{CaCO_3}=\dfrac{400}{100}=4\left(mol\right)\)
`=>` \(n_{CaCO_3\left(pư\right)}=4.75\%=3\left(mol\right)\)
PTHH: \(CaCO_3\xrightarrow[]{t^o}CaO+CO_2\)
3--------->3-------->3
`=>` \(m_{CO_2}=3.44=132\left(g\right)\)
Theo ĐLBTKL: \(m_X=m_{đá.vôi}-m_Y=500-132=368\left(g\right)\)
b) \(m_{CaO}=3.56=168\left(g\right)\)
c) Ta có: \(n_{NaOH}=800.2\%=16\left(g\right)\Rightarrow n_{NaOH}=\dfrac{16}{40}=0,4\left(mol\right)\)
Xét \(T=\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,4}{3}< 1\Rightarrow\) Phản ứng tạo muối axi, CO2 dư
PTHH: \(NaOH+CO_2\rightarrow NaHCO_3\)
0,4------>0,4------>0,4
`=>` \(m_{dd}=0,4.44+800=817,6\left(g\right)\)
`=>` \(C\%_{NaHCO_3}=\dfrac{0,4.84}{817,6}.100\%=4,11\%\)