Question

Giúp vs

Answer 1

\(n_{CO}=\dfrac{v}{22,4}=\dfrac{15,6}{22,4}\approx0,7mol\)

CuO+CO\(\rightarrow\)Cu+CO₂

x........x

Fe₂O₃+3CO\(\rightarrow\)2Fe+3CO₂

y...........3y

-Ta có hệ: \(\left\{{}\begin{matrix}80x+160y=40\\x+3y=0,7\end{matrix}\right.\)

Giải ra x=0,1 và y=0,2

\(m_{CuO}=0,1.80=8gam\)

\(m_{Fe_2O_3}=0,2.160=32gam\)

%CuO=\(\dfrac{8.100}{40}=20\%\)

%Fe₂O₃=80%

Answer 2

Ba(OH)₂+2HCl\(\rightarrow\)BaCl₂+2H₂O

x..............2x...........x..........2x

NaOH+HCl\(\rightarrow\)NaCl+H₂O

y..........y...........y........y

-Ta có hệ: \(\left\{{}\begin{matrix}171x+40y=42,2\\208x+58,5y=53,3\end{matrix}\right.\)

Giải ra x=0,2 và y=0,2

\(m_{Ba\left(OH\right)_2}=171.0,2=34,2gam\)

\(m_{NaOH}=40.0,2=8gam\)

%Ba(OH)₂=\(\dfrac{34,2.100}{42,2}\approx81\%\)

%NaOH=19%

\(n_{HCl}=2x+y=2.0,2+0,2=0,6mol\)

\(m_{dd_{HCl}}=\dfrac{0,6.36,5.100}{7,3}=300gam\)

\(m_{dd}=42,2+300=342,2gam\)

C%BaCl₂=\(\dfrac{208.0,2.100}{342,2}\approx12,2\%\)

C%NaCl=\(\dfrac{0,2.58,5.100}{342,2}\approx3,42\%\)