a, \(n_{H_2}=\dfrac{22,4}{22,4}=1\left(mol\right)\)

PT: \(2Na+2H_2O\rightarrow2NaOH+H_2\)

Theo PT: \(n_{Na}=n_{NaOH}=2n_{H_2}=2\left(mol\right)\)

\(\Rightarrow m_{Na}=2.23=46\left(g\right)\)

b, \(m_{NaOH}=2.40=80\left(g\right)\)

c, \(n_{HCl}=\dfrac{365}{36,5}=10\left(mol\right)\)

PT: \(NaOH+HCl\rightarrow NaCl+H_2O\)

Xét tỉ lệ: \(\dfrac{2}{1}< \dfrac{10}{1}\), ta được HCl dư.

Theo PT: \(n_{NaCl}=n_{NaOH}=2\left(mol\right)\Rightarrow m_{NaCl}=2.58,5=117\left(g\right)\)

d, \(n_{H_2}=\dfrac{1}{3}\left(mol\right)\)

- Với Fe₃O₄:

PT: \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)

Theo PT: \(n_{Fe}=\dfrac{3}{4}n_{H_2}=0,25\left(mol\right)\Rightarrow m_{Fe}=0,25.56=14\left(g\right)\)

- Với CuO:

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

Theo PT: \(n_{Cu}=n_{H_2}=\dfrac{1}{3}\left(mol\right)\Rightarrow m_{Cu}=\dfrac{1}{3}.64=\dfrac{64}{3}\left(g\right)\)

giúp e với mọi người

Giả sử CTHH của oxit cần tìm là A₂O_n

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

PT: \(A_2O_n+nH_2\underrightarrow{t^o}2A+nH_2O\)

Theo PT: \(n_{A_2O_n}=\dfrac{1}{n}n_{H_2}=\dfrac{0,15}{n}\left(mol\right)\)

\(\Rightarrow M_{A_2O_n}=\dfrac{8}{\dfrac{0,15}{n}}=\dfrac{160}{3}n\left(g/mol\right)\)

\(\Rightarrow2M_A+16n=\dfrac{160}{3}n\Rightarrow M_A=\dfrac{56}{3}\left(g/mol\right)\)

Với n = 3, M_A = 56 (g/mol) là thỏa mãn.

Vậy: KL đó là Fe.

a, \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

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

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,15\left(mol\right)\Rightarrow V_{H_2}=0,15.22,4=3,36\left(l\right)\)

b, Có lẽ đề hỏi bao nhiêu gam đồng thay vì "bao nhiêu gam sắt" bạn nhỉ?

\(n_{CuO}=\dfrac{20}{80}=0,25\left(mol\right)\)

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

Xét tỉ lệ: \(\dfrac{0,25}{1}>\dfrac{0,15}{1}\), ta được CuO dư.

Theo PT: \(n_{Cu}=n_{H_2}=0,15\left(mol\right)\Rightarrow m_{Cu}=0,15.64=9,6\left(g\right)\)

Gọi CTPT của A là C_nH_2n+2O

Mà: %O = 34,78%

\(\Rightarrow\dfrac{16}{12n+2n+2+16}=0,3478\Rightarrow n=2\)

Vậy: CTPT của A là C₂H₆O

Đáp án: A

a, \(Na_2O+H_2O\rightarrow2NaOH\)

b, \(n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\)

Theo PT: \(n_{NaOH}=2n_{Na_2O}=0,4\left(mol\right)\Rightarrow m_{NaOH}=0,4.40=16\left(g\right)\)

c, \(C\%_{NaOH}=\dfrac{16}{200}.100\%=8\%\)

a, \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

b, \(n_{Fe}=\dfrac{33,6}{56}=0,6\left(mol\right)\)

Theo PT: \(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe}=0,3\left(mol\right)\Rightarrow m_{Fe_2O_3}=0,3.160=48\left(g\right)\)

c, \(n_{H_2}=\dfrac{3}{2}n_{Fe}=0,9\left(mol\right)\Rightarrow V_{H_2}=0,9.22,4=20,16\left(l\right)\)

thanks 😙👍

\(n_{C_2H_5OH}=\dfrac{4,6}{46}=0,1\left(mol\right)\)

PT: \(2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\)

Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{C_2H_5OH}=0,05\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,05.22,4=1,12\left(l\right)\)

Đáp án: D

\(n_{C_2H_5OH}=\dfrac{4,6}{23}=0,2\left(mol\right)\\ C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\\ n_{H_2}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ V=V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ Chọn.A\)

\(n_K=\dfrac{39}{39}=1\left(mol\right)\\ 2K+2H_2O\rightarrow2KOH+H_2\\ n_{H_2}=\dfrac{1}{2}=0,5\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,5.22,4=11,2\left(l\right)\\ b,n_{KOH}=n_K=1\left(mol\right)\\ C_{MddKOH}=\dfrac{1}{0,2}=5\left(M\right)\\ c,2H_2+O_2\rightarrow\left(t^o\right)2H_2O\\ n_{O_2}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\)

Bài 2:

\(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\\a, 2Na+2H_2O\rightarrow2NaOH+H_2\\ b,n_{H_2}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ c,NaOH:Tính.bazo\Rightarrow Quỳ.tím.hoá.xanh\)

Câu 3: Em chụp lại bảng nha