\(n_{SO_2}=\dfrac{44,8}{22,4}=2\left(mol\right)=>m_{SO_2}=2.64=128\left(g\right)\)

\(n_{CO}=\dfrac{5,6}{22,4}=0,25\left(mol\right)=>m_{CO}=0,25.28=7\left(g\right)\)

=> m_hh = 128 + 7 = 135 (g)

=> A

A

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

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

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

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

\(\Rightarrow A\)

\(n_{CO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\\ n_{CaCO_3}=n_{CO_2}=0,05\left(mol\right)\\ m_{rắn}=m_{CaCO_3}=0,05.100=5\left(g\right)\\ \Rightarrow ChọnD\)

Ta có: \(n_{CO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Bảo toàn Cacbon: \(n_{CO_2}=n_{CaCO_3}=0,05\left(mol\right)\) \(\Rightarrow m_{CaCO_3}=0,05\cdot100=5\left(g\right)\)

\(\Rightarrow\) Chọn D

\(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)

Theo PT: \(n_{H_2O}=n_{H_2}=0,6\left(mol\right)\Rightarrow m_{H_2O}=0,6.18=10,8\left(g\right)\)

→ 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: $n_{Ba}=n_{H_2}=0,25(mol)$

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

$\to A$ 

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

=> 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

B A B A

Gọi x,y lần lượt là số mol của Al, Mg

n_H2 =