\(Fe+2HCl\rightarrow FeCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Đặt:\left\{{}\begin{matrix}n_{Fe}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}56x+27y=2,78\\x+1,5y=0,07\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,04\\y=0,02\end{matrix}\right.\\ n_{FeCl_2}=n_{Fe}=0,04\left(mol\right)\\ \Rightarrow m_{FeCl_2}=0,04.127=5,08\left(g\right)\)

\(n_{H_2}=\frac{5,6}{22,4}=0,25(mol)\\ m_{H_2}=0,25.2=0,5(g)\\ BT H:\\ n_{HCl}=2n_{H_2}=0,25.2=0,5(mol)\\ m_{HCl}=0,5.36,5=18,25(g)\\ BTKL:\\ m_{hh}+m_{HCl}=m_{muối}+m_{H_2}\\ 15+18,25=m_{muối}+0,5\\ \to m_{muối}=32,75(g)\\ \to D\)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(n_{Fe}=n_{H_2}=0,15\left(mol\right)\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)

Đáp án: B

^{\(n_{H_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)}

^{\(Ba+2HCl\rightarrow BaCl_2+H_2\)}

^{\(0.25................................0.25\)}

\(m_{Ba}=0.25\cdot137=34.25\left(g\right)\)

$Ba+2HCl\to BaCl_2+H_2\uparrow$

$n_{H_2}=\dfrac{5,6}{22,4}=0,25(mol)$

Theo PT: $n_{Ba}=n_{H_2}=0,25(mol)$

$\Rightarrow m_{Ba}=0,25.137=34,25(g)$

$\to A$ 

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

0,1                                         0,1   ( mol )

\(m_{Fe}=0,1.56=5,6g\)

\(\rightarrow m_{Cu}=10-5,6=4,4g\)

--> B

B

PTHH: \(CuO+CO\xrightarrow[]{t^o}Cu+CO_2\)

            \(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)

Ta có: \(n_{CuO}=\dfrac{6}{80}=0,075\left(mol\right)=n_{CO_2}=n_{CaCO_3}\)

\(\Rightarrow m_{CaCO_3}=100\cdot0,075=7,5\left(g\right)\)

\(PTHH:H_2+CuO\underrightarrow{t^o}Cu+H_2O\)

Áp dụng định luật bảo toàn khối lượng, ta có:

\(m_{H_2}+m_{CuO}=m_{Cu}+m_{H_2O}\\ m_{H_2O}=m_{H_2}+m_{CuO}-m_{Cu}=2+80-64=18\left(g\right)\\ \Rightarrow D\)

Ta có PTHH: \(3H_2+Fe_2O_3\underrightarrow{t\text{°}}2Fe+3H_2O\)

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

- Theo PTHH:

 + Để thu được \(\text{2 mol Fe}\) cần đun nóng \(\text{1 mol Fe}_2O_3\)

 ⇒ Để thu được \(\text{0,1 mol Fe}\) cần đun nóng \(\text{0,05 mol Fe}_2O_3\)

\(m_{Fe_2O_3}=0,05\text{ x }160=8\left(g\right)\)

Vậy:Khối lượng Sắt (III) oxit đã tham gia phản ứng là 8g

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