Xăng, dầu đk nhỉ

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

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

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

Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT: \(n_{Fe}=n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Cu}=24-m_{Fe}=12,8\left(g\right)\) \(\Rightarrow n_{Cu}=\dfrac{12,8}{64}=0,2\left(mol\right)\)

Theo PT: \(\left\{{}\begin{matrix}n_{CuO}=n_{Cu}=0,2\left(mol\right)\\n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe}=0,1\left(mol\right)\end{matrix}\right.\)

⇒ m = m_CuO + m_Fe2O3 = 0,2.80 + 0,1.160 = 32 (g)

b, \(\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{0,2.80}{32}.100\%=50\%\\\%m_{Fe_2O_3}=50\%\end{matrix}\right.\)

c, Theo PT: \(n_{H_2}=n_{CuO}+3n_{Fe_2O_3}=0,5\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,5.22,4=11,2\left(l\right)\)

a, PT: \(2Na+2H_2O\rightarrow2NaOH+H_2\)

\(Na_2O+H_2O\rightarrow2NaOH\)

Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Theo PT: \(n_{Na}=2n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Na}=0,2.23=4,6\left(g\right)\)

\(\Rightarrow m_{Na_2O}=23,2-m_{Na}=18,6\left(g\right)\)

b, - Dd A làm quỳ tím chuyển xanh vì A chứa bazo tan.

Ta có: \(n_{Na_2O}=\dfrac{18,6}{62}=0,3\left(mol\right)\)

Theo PT: \(n_{NaOH}=n_{Na}+2n_{Na_2O}=0,8\left(mol\right)\)

\(\Rightarrow m_{NaOH}=0,8.40=32\left(g\right)\)

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

Ta có: \(n_{CuO}=\dfrac{6,4}{80}=0,08\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,08}{1}< \dfrac{0,1}{1}\), ta được H₂ dư.

Gọi n_{CuO (pư)} = a (mol) ⇒ n_{CuO (dư)} = 0,08 - a (mol)

Theo PT: \(n_{Cu}=n_{CuO\left(pư\right)}=a\left(mol\right)\)

⇒ m _{chất rắn} = m_Cu + m_{CuO (dư)} = 64a + 80.(0,08-a) = 5,44

⇒ a = 0,06 (mol)

\(\Rightarrow H\%=\dfrac{0,06}{0,08}.100\%=75\%\)

a, PT: \(2Mg+O_2\underrightarrow{t^o}2MgO\)

\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

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

b, Ta có: \(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Theo ĐLBT KL, có: m _oxit = m_KL + m_O2 = 15,6 + 0,2.32 = 22 (g)

c, Gọi: \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\) (trong 15,6 g)

⇒ 24x + 27y = 15,6 (1)

Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{Mg}+\dfrac{3}{4}n_{Al}=\dfrac{1}{2}x+\dfrac{3}{4}y=0,2\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=1,4\\y=-\dfrac{2}{3}\end{matrix}\right.\)

Đến đây thì ra số mol âm, bạn xem lại đề nhé.

\(a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ b.m_{Fe}=60,5.66,77\%=41g\\ \Rightarrow n_{Fe}=\dfrac{41}{56}=0,73\left(mol\right)\\ n_{Zn}=\dfrac{60,5-41}{65}=0,3\left(mol\right)\\ \Sigma n_{H_2}=n_{Fe}+n_{Zn}=0,73+0,3=1,03\left(mol\right)\\ V_{H_2}=1,03.22,4=23,072\left(l\right)\)

a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

b, Ta có: \(m_{Fe}=60,5.67,77\%=41\left(g\right)\Rightarrow n_{Fe}=\dfrac{41}{56}\left(mol\right)\)

\(m_{Zn}=60,5-41=19,5\left(g\right)\Rightarrow n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Zn}+n_{Fe}=\dfrac{289}{280}\left(mol\right)\)

\(\Rightarrow V_{H_2}=\dfrac{289}{280}.22,4=23,12\left(l\right)\)

\(V_{H_2}\) là 19,6l mà anh

\(\left\{{}\begin{matrix}m_{CuO}=50.20\%=10\left(g\right)\\m_{Fe_2O_3}=50-10=40\left(g\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n_{CuO}=\dfrac{10}{80}=0,125\left(mol\right)\\n_{Fe_2O_3}=\dfrac{40}{160}=0,25\left(mol\right)\end{matrix}\right.\)

PTHH:
\(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)

0,125->0,125

\(Fe_2O_3+3H_2\xrightarrow[]{t^o}2Fe+3H_2\)

0,25--->0,75

\(\Rightarrow V_{H_2}=\left(0,75+0,125\right).22,4=18,2\left(l\right)\)

a) \(n_{Fe}=\dfrac{12}{56}=\dfrac{3}{14}\left(mol\right)\)

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

            \(\dfrac{3}{14}\)---------------------->\(\dfrac{3}{14}\)

\(\Rightarrow V_{H_2}=\dfrac{3}{14}.22,4=4,8\left(l\right)\)

b) \(n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\)

PTHH: \(ZnO+H_2\xrightarrow[]{t^o}Zn+H_2O\)

Xét tỉ lệ: \(0,1< \dfrac{3}{14}\Rightarrow H_2\) dư

Theo PT: \(n_{Zn}=n_{ZnO}=0,1\left(mol\right)\Rightarrow m_{Zn}=0,1.65=6,5\left(g\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(1mol\)                             \(1mol\)

\(\dfrac{3}{14}mol\)                        \(\dfrac{3}{14}mol\)

\(a)n_{Fe}=\dfrac{m}{M}=\dfrac{12}{56}\approx0,21=\dfrac{3}{14}\left(mol\right)\)

\(V_{H_2}=n.22,4=\dfrac{3}{14}.22,4=4,8\left(l\right)\)

\(b)n_{ZnO}=\dfrac{m}{M}=\dfrac{8,1}{81}=0,1\left(mol\right)\)

\(ZnO+H_2\rightarrow Zn+H_2O\)

\(1mol\)    \(1mol\)    \(1mol\)

\(0,1mol\)  \(0,1mol\)  \(0,1mol\)

\(\text{Ta thấy }H_2\text{ dư,ZnO phản ứng hết.Bài toán tính theo ZnO}\)

\(m_{Zn}=n.M=0,1.65=6,5\left(g\right)\)

\(M_A=5.2=10\left(g/mol\right)\)

Do các khí đo ở cùng điều kiện nhiệt độ và áp suất nên tỉ lệ thể tích cũng là tỉ lệ mol

Chọn \(\left\{{}\begin{matrix}n_{H_2}=17,5\left(mol\right)\\n_{N_2}=5\left(mol\right)\end{matrix}\right.\)

Gọi \(n_{N_2\left(p\text{ư}\right)}=a\left(mol\right)\left(0< a< 5\right)\)

PTHH: \(N_2+3H_2\xrightarrow[]{t^o,p,xt}2NH_3\)

             a---->3a---------->2a

Xét tỉ lệ: \(5< \dfrac{17,5}{3}\Rightarrow\) Hiệu suất phản ứng tính theo N₂

Ta có: \(n_A=5+17,5+2a-a-3a=22,5-2a\left(mol\right)\)

Theo ĐLBTKL: \(m_A=5.28+17,5.2=175\left(g\right)\)

\(\Rightarrow M_A=\dfrac{175}{22,5-2a}=10\Leftrightarrow a=2,5\left(TM\right)\)

\(\Rightarrow H=\dfrac{2,5}{5}.100\%=50\%\)

a, PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

b, Ta có: \(n_{P_2O_5}=\dfrac{7,1}{142}=0,05\left(mol\right)\)

Theo PT: \(n_P=2n_{P_2O_5}=0,1\left(mol\right)\)

\(\Rightarrow m_P=0,1.31=3,1\left(g\right)\)

\(n_{O_2}=\dfrac{5}{2}n_{P_2O_5}=0,125\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,125.22,4=2,8\left(l\right)\)

c, Có: \(V_{O_2\left(dư\right)}=2,8.15\%=0,42\left(l\right)\)

\(\Rightarrow V_{O_2}=2,8+0,42=3,22\left(l\right)\)