D

\(n_{O_2}=\dfrac{44,8}{22,4}=2\left(mol\right)\)

\(\Rightarrow m_{O_2}=32.2=64\left(g\right)\)

Chọn D

Chọn D nha

\(n_{CO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

=> n_C = 0,4 (mol)

\(n_{H_2O}=\dfrac{9}{18}=0,5\left(mol\right)\)

=> n_H = 1 (mol)

m = m_C + m_H = 0,4.12 + 1.1 = 5,8 (g)

=> A

\(n_{CO_2}=\dfrac{44.8}{22.4}=2\left(mol\right)\)

\(C+O_2\underrightarrow{^{^{t^0}}}CO_2\)

\(2..............2\)

\(m_C=2\cdot12=24\left(g\right)\)

\(\Rightarrow A\)

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

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

0,1                                         0,1   ( mol )

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

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

--> B

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

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

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

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

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

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

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

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

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

$\to A$ 

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

Đặt x là số mol Fe => 2x là số mol Mg

PTHH: Fe +2 HCl -> FeCl2 + H2

Mg + 2 HCl -> MgCl2 + H2

Ta có: nH2(tổng)=nFe+nMg=3x(mol)

<=> 0,3=3x

<=>x=0,1

=> m(muối)= mFeCl2 + mMgCl2 = 0,1. 127+ 0,2.95= 31,7(g)

=> CHỌN A