Ta có: \(\left\{{}\begin{matrix}n_{Ba^{2+}}=n_{Ba\left(OH\right)_2}=0,1.0,1=0,01\left(mol\right)\\n_{SO_4^{2-}}=n_{H_2SO_4}=0,1.0,15=0,015\left(mol\right)\end{matrix}\right.\)

PT ion rút gọn:        \(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\downarrow\)

Ban đầu:                0,01       0,015

Pư:                         0,01---->0,01

Sau pư:                   0           0,005        0,01

=> \(m_X=m_{BaSO_4}=0,01.233=2,33\left(g\right)\)

Parafin + Khí oxi -> khí cacbon đioxit + hơi nước

\(Para\text{f}in+Oxi\rightarrow Cacbon\text{ }ddioxit+Hơi\text{ }nước\)

a) PTHH: \(Fe_2O_3+3CO\xrightarrow[]{t^o}2Fe+3CO_2\)

b) Theo ĐLBTKL:

\(m_{Fe_2O_3}+m_{CO}=m_{Fe}+m_{CO_2}\)

=> \(a=m_{CO_2}=8,4+16-11,2=13,2\left(g\right)\)

c) \(n_{CO}=\dfrac{m}{M}=\dfrac{8,4}{28}=0,3\left(mol\right)\)

d) \(n_{CO_2}=\dfrac{m}{M}=\dfrac{13,2}{44}=0,3\left(mol\right)\)

=> \(V_{CO_2\left(đktc\right)}=n.22,4=0,3.22,4=6,72\left(l\right)\)

Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right);n_{NO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PTHH:

Phần 1: \(Cu+4HNO_{3\left(đặc\right)}\rightarrow Cu\left(NO_3\right)_2+2NO_2\uparrow+2H_2O\)

             0,2<-------------------------------------0,4

Phần 2: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

              0,2<--------------------------0,3

b) \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.2.27}{0,2.2.27+0,2.2.64}.100\%=29,67\%\\\%m_{Cu}=100\%-29,67\%=70,33\%\end{matrix}\right.\)

Nguyên tố M gồm các hạt: p, e, n

Nguyên tố R gồm các hạt : p', e', n'

Ta có: \(\left\{{}\begin{matrix}p+2p'=96\\\%M=\dfrac{p+n}{\left(p+n\right)+2\left(p'+n'\right)}=25,926\%\end{matrix}\right.\\\Leftrightarrow \left\{{}\begin{matrix}p+2p'=96\\\%M=\dfrac{p+p+4}{\left(p+p+4\right)+2\left(p'+p'+10\right)}=25,926\%\end{matrix}\right.\\\Leftrightarrow\left\{{}\begin{matrix}p+2p'=96\\\%M=\dfrac{2p+4}{2p+4p'+24}=25,926\%\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}p=26\\p'=35\end{matrix}\right.\)

=> M là Fe, R là Brom

Vậy X là \(FeBr_2\)

\(n_{O_2}=\dfrac{20,16}{22,4}=0,9\left(mol\right)\)

PTHH: \(C+O_2\xrightarrow[]{t^o}CO_2\)

          0,9<-0,9

=> \(m_C=0,9.12=10,8\left(g\right)\)

=> \(m=m_{than.đá}=\dfrac{10,8}{\left(100-4\right)\%}=11,25\left(g\right)\)