\(n_{Co_2}=\dfrac{3,36}{22,4}=0,15mol\)

\(n_{N_2}=\dfrac{44,8}{22,4}=2mol\)

Chào Hoàng lâu ko gặp

\(n_{CO_2}=\dfrac{V_{CO_2}}{22,4}=\dfrac{44,8}{22,4}=2\left(mol\right)\)

2 mol

+ \(M_{CO_2}=12+16.2=44\left(\dfrac{g}{mol}\right)\)

\(n_{CO_2}=\dfrac{m}{M}=\dfrac{11}{44}=0,25mol\)

+ \(n_{H_2}=\dfrac{9.10^{23}}{6.10^{23}}=1,5mol\)

\(V_{H_2\left(đktc\right)}=n.22,4=1,5.22,4=33,6\left(l\right)\)

Câu 2

\(n_{O_2}=\dfrac{V_{O_2\left(đktc\right)}}{22,4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Số phân tử oxi \(=\)\(0,3.6.10^{23}=1,8.10^{23}\)

\(M_{O_2}=16.2=32\left(\dfrac{g}{mol}\right)\)

\(m_{O_2}=n.M=0,3.32=9,6g\)

nO2 = 44,8 : 22,4 = 2 (l) 
pthh X + O2 -->2 CO2 +H2O 
                 2---> 4-------> 2 (mol) 
=> mCO2 = 4 . 44 = 176(g) 
=> mH2O = 2.18 = 36 (g)

tỉ lệ là 2:1 hay 1:2

-Ta có:

\(m_{CO_2}=16+12,2=44\)(g)

\(n_{CO_2}=\dfrac{22}{44}=0,5\left(mol\right)\)

-\(n_{H_2}=\dfrac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\)

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

n=5,6/22,4=0,25(mol)

m=0,25*44=11(g)

Đặt \(n_{O_2}=x;n_{CO_2}=y\)

\(n_X=\dfrac{4,48}{22,4}=0,2\left(mol\right)\Leftrightarrow x+y=0,2\) 

Ta có: \(16x+44y=\left(x+y\right).18.2\)

     \(\Leftrightarrow2y=5x\)

     \(\Leftrightarrow\dfrac{y}{5}=\dfrac{x}{2}\)

Mà x+y=0,2

\(\Rightarrow\dfrac{y}{5}=\dfrac{x}{2}=\dfrac{x+y}{5+2}=\dfrac{0,2}{7}=0,0286\)

\(\Rightarrow y=5.0,0286=0,143\left(mol\right);x=0,2-0,143=0,057\left(mol\right)\)

Ta có: \(n_X=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

Gọi: \(\left\{{}\begin{matrix}n_{H_2}=x\left(mol\right)\\n_{CO_2}=y\left(mol\right)\end{matrix}\right.\) ⇒ x + y = 0,4 (mol) (1)

\(d_{X/NO_2}=0,5\Rightarrow M_X=0,5.46=23\left(g/mol\right)\)

⇒ m_X = 0,4.23 = 9,2 (g)

⇒ 2x + 44y = 9,2 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,2\left(mol\right)\\y=0,2\left(mol\right)\end{matrix}\right.\)

a.

\(m_{Al}=0.5\cdot27=13.5\left(g\right)\)

\(m_{CO_2}=\dfrac{6.72}{22.4}\cdot44=13.2\left(g\right)\)

\(m_{N_2}=\dfrac{5.6}{22.4}\cdot28=7\left(g\right)\)

\(m_{CaCO_3}=0.25\cdot100=25\left(g\right)\)

b.

\(m_{hh}=\dfrac{3.36}{22.4}\cdot2+\dfrac{5.6}{22.4}\cdot28+0.2\cdot44=16.1\left(g\right)\)

e cảm ơn^^

\(m_{CO_2}=12+16\cdot2=44\)

Số mol của CO₂

\(n=\frac{m}{M}=\frac{11}{44}=0,25\)(mol)

Số mol của H₂

\(n=\frac{sophantu}{6.10^{23}}=\frac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\)

\(\Rightarrow\) V_H2(đktc) =n.22,4=1,5.22,4=33,6(lít)

• n_CO2 = 11 / 44 = 0,25 (mol)
• n_H2 = \(\frac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\)

=> V_H2(đktc) = 1,5 x 22,4 = 33,6 lít