\(a,m_{H_2}=M_{H_2}.n_{H_2}=2.0,75=1,5\left(g\right)\\ b,m_{NO}=n_{NO}.M_{NO}=0,5.30=15\left(g\right)\\ c,m_{Cl_2}=M_{Cl_2}.n_{Cl_2}=71.1,5=106,5\left(g\right)\\ d,m_{CO}=n_{CO}.M_{CO}=0,25.28=7\left(g\right)\\ e,m_{SO_2}=M_{SO_2}.n_{SO_2}=3.64=192\left(g\right)\\ f,m_{O_2}=n_{O_2}.M_{O_2}=0,5.32=16\left(g\right)\\ g,m_{NO_2}=M_{NO_2}.n_{NO_2}=2,5.46=115\left(g\right)\\ h,m_{H_2S}=n_{H_2S}.M_{H_2S}=0,25.34=8,5\left(g\right)\\ i,m_{HCl}=M_{HCl}.n_{HCl}=36,5.1=36,5\left(g\right)\\ j,m_{NH_3}=n_{NH_3}.M_{NH_3}=1,25.17=21,25\left(g\right)\)

\(n_{O_2}=\dfrac{22,4}{22,4}=1\left(mol\right);n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ 4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)

a) Đây là phản ứng hoá hợp vì từ 2 chất phản ứng nhưng chỉ tạo thành duy nhất 1 chất sản phẩm.

\(b,Vì:\dfrac{0,2}{4}< \dfrac{1}{3}\Rightarrow O_2dư\\ n_{O_2\left(dư\right)}=1-\dfrac{3}{4}.0,2=0,85\left(mol\right)\\ \Rightarrow m_{O_2\left(dư\right)}=0,85.32=27,2\left(g\right)\)

\(c,n_{Al_2O_3}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ \Rightarrow m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)

a. \(n_{NaOH\left(5\%\right)}=\frac{100.5}{100.40}=0,125mol\)

\(m_{ddNaOH\left(8\%\right)}=\frac{0,125.100.40}{8}=62,5g\)

\(m_{H_2O}\) bay hơi \(=100-62,5=37,5g\)

b. \(m_{NaOH\left(A\right)}=40.0,125=5g\)

\(m_{NaOH\left(8\%\right)}=\frac{100.8}{100}=8g\)

\(m_{NaOH}\) thêm vào \(=8-5=3g\)

Câu 5:

\(a,Đặt:n_{Cu}=a\left(mol\right);n_{Fe}=b\left(mol\right)\\ 2Cu+O_2\rightarrow\left(t^o\right)2CuO\\ 3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\\ \Rightarrow\left\{{}\begin{matrix}11,2a+\dfrac{44,8}{3}b=4,48\\64a+56b=21,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,15\end{matrix}\right.\\ \Rightarrow m_{Cu}=64a=12,8\left(g\right);m_{Fe}=56b=8,4\left(g\right)\)

\(b,CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\\ n_{CH_4}=\dfrac{1,6}{16}=0,1\left(mol\right);n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ Vì:\dfrac{0,1}{1}< \dfrac{0,3}{2}\Rightarrow O_2dư\\ \Rightarrow n_{O_2\left(dư\right)}=0,3-0,1.2=0,1\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)

Câu 4:

\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\left(1\right)\\ 2Mg+O_2\rightarrow\left(t^o\right)2MgO\left(2\right)\\ n_{O_2\left(2\right)}=\dfrac{n_{Mg}}{2}=\dfrac{4,8:24}{2}=0,1\left(mol\right)\\ \Rightarrow n_{O_2\left(1\right)}=\dfrac{6,72}{22,4}-0,1=0,2\left(mol\right)\\ n_{Fe}=\dfrac{3}{2}.n_{O_2\left(1\right)}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\\ \Rightarrow a=m_{Fe}+m_{Mg}=0,3.56+4,8=21,6\left(g\right)\)

a) PTHH:

\(4CO+Fe_3O_4\rightarrow3Fe+4CO_2\left(1\right)\)

\(3H_2+Fe_2O_3\rightarrow2Fe+3H_2O\left(2\right)\)

b) Theo PTPƯ trên ta có:

Muốn khử 1 mol \(Fe_3O_4\) cần 4 mol CO

Muốn khử 0,2 mol \(Fe_3O_4\) cần x mol CO

\(\Rightarrow x=0,2.4=0,8\left(mol\right)\)

\(V_{CO}=0,8.22,4=17,92\left(l\right)\)

Muốn khử 1 mol \(Fe_2O_3\) cần 3 mol H2

Muốn khử 0,2 mol \(Fe_2O_3\) cần y mol H2

\(\Rightarrow y=0,2.3=0,6\left(mol\right)\)

\(V_{H_2}=0,6.22,4=13,44\left(l\right)\)

c) (1): \(n_{Fe}=3n_{Fe_3O_4}=3.0,2=0,6\left(mol\right)\)

\(m_{Fe}=0,6.56=33,6\left(g\right)\)

(2): \(n_{Fe}=2n_{Fe_2O_3}=2.0,2=0,4\left(mol\right)\)

\(m_{Fe}=0,4.56=22,4\left(g\right)\)

\(n_{Cl_2}=\dfrac{0,56}{22,4}=0,025\left(mol\right)\)

PTHH: 2X + nCl₂ --t^o--> 2XCl_n

       \(\dfrac{0,05}{n}\)<-0,025

=> \(M_X=\dfrac{1,15}{\dfrac{0,05}{n}}=23n\left(g/mol\right)\)
Xét n = 1 => M_X = 23(Na)

Em cảm ơn ạ!

\(C+O_2\rightarrow\left(t^o\right)CO_2\\ S+O_2\rightarrow\left(t^o\right)SO_2\\ 4K+O_2\rightarrow2K_2O\\ 3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\\ 4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\\ 2Mg+O_2\rightarrow\left(t^o\right)2MgO\\ 2C_4H_{10}+13O_2\rightarrow\left(t^o\right)8CO_2+10H_2O\)

N nguyên tử thường không phản ứng oxi. Mà Nito phân tử mới phản ứng với oxi.

N không có tồn tại dạng nguyên tử mà phản ứng với oxi như thế đâu em

Có \(\left\{{}\begin{matrix}4.p_M+2n_M+2p_O+n_O=140\\4.p_M+2p_O-2n_M-n_O=44\end{matrix}\right.\)

=> \(2p_M+p_O=46\)

=> 2.p_M + 8 = 46

=> p_M = 19

Vậy M là K

Sai rồi nhé

8 proton , 8 electron , 8 notron