a, khối lượng của 2,5 mol CuO là:
\(m=n.M=2,5.80=200\left(g\right)\)

b, số mol của 4,48 lít khí CO2 (đktc) là:
\(n=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

a.

mCuO=n.M=2,5.(1.64+1.16)= 200 mol

b.nCO₂=V/22,4=4,48:22,4=0,2 mol

a) PTHH: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

Ta có: \(n_{C_2H_4}=\dfrac{5,6}{28}=0,2\left(mol\right)=n_{C_2H_4Br_2}\) \(\Rightarrow m_{C_2H_4Br_2}=0,2\cdot188=37,6\left(g\right)\)

b) Ta có: \(n_{CH_4}=\dfrac{11,2}{22,4}-0,2=0,3\left(mol\right)\)

Bảo toàn nguyên tố: \(n_{CO_2}=n_{CH_4}+2n_{C_2H_4}=0,7\left(mol\right)\)

\(\Rightarrow V_{CO_2}=0,7\cdot22,4=15,68\left(l\right)\)

\(n_C = n_{CO_2}= \dfrac{11,2}{22,4} = 0,5(mol)\\ n_H = 2n_{H_2O} = 2.\dfrac{9}{18} = 1(mol)\\ n_A = \dfrac{11,2}{22,4} = 0,5(mol)\\ M_A = \dfrac{1,34}{\dfrac{1}{22,4}}=30\)

Số nguyên tử Cacbon = \(\dfrac{n_C}{n_A} = \dfrac{0,5}{0,5} = 1\)

Số nguyên tử Hidro = \(\dfrac{n_H}{n_A} = \dfrac{1}{0,5} = 2\)

Số nguyên tử Oxi = \(\dfrac{30-12-2}{16}=1\)

Vậy CTPT của A :CH₂O

\(a)n_{O_2}=\dfrac{V}{24,79}=\dfrac{11,2}{24,79}\approx0,45\left(mol\right)\\ \Rightarrow m_{O_2}=n.M=0,45.\left(16.2\right)=14,4\left(g\right)\\ b)n_{NaCl}=\dfrac{V}{24,79}=\dfrac{12,5}{24,79}\approx0,505\left(mol\right)\\ \Rightarrow m_{NaCl}=n.M=0,505.\left(23+35,5\right)=29,5425\left(g\right).\)

\(n_{O_2}=\dfrac{11.2}{22.4}=0.5\left(mol\right)\\ m_{O_2}=n\cdot M=0.5\cdot32=16\left(gam\right)\)

\(n_{NaCl}=\dfrac{12.5}{22.4}\approx0.5\left(mol\right)\\ m_{NaCl}=n\cdot M=0.5\cdot\text{58.5}=29.25\left(gam\right)\)

Bài 1:

PTHH: \(2C_4H_{10}+13O_2\xrightarrow[]{t^o}8CO_2+10H_2O\)

Ta có: \(n_{C_4H_{10}}=\dfrac{11,6}{58}=0,2\left(mol\right)\) 

\(\Rightarrow\left\{{}\begin{matrix}n_{CO_2}=0,8\left(mol\right)\\n_{H_2O}=1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CO_2}=0,8\cdot44=35,2\left(g\right)\\m_{H_2O}=1\cdot18=18\left(g\right)\end{matrix}\right.\)

Bài 2:

PTHH: \(CaCO_3\xrightarrow[]{t^o}CaO+CO_2\uparrow\)

Ta có: \(n_{CO_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)=n_{CaO}=n_{CaCO_3\left(p.ứ\right)}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CaO}=0,5\cdot56=28\left(g\right)\\\%m_{CaCO_3\left(p.ứ\right)}=\dfrac{0,5\cdot100}{100}\cdot100\%=50\%\end{matrix}\right.\)

Bài 3:

PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

Tính theo sản phẩm

Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)=n_{MgCl_2}\) \(\Rightarrow m_{MgCl_2}=0,2\cdot95=19\left(g\right)\)

nNH_3(đktc) =\(\dfrac{V}{22,4}\)=\(\dfrac{11,2}{22,4}\)= 0.5 mol 

=>mNH₃ = n.M = 0,5 . 17= 8,5 gam

\(n_{H_2}=\dfrac{22,4}{22,4}=1\left(mol\right)=>m_{H_2}=1.2=2\left(g\right)\)

\(n_{N_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)=>m_{H_2}=0,15.28=4,2\left(g\right)\)

\(n_{CO}=\dfrac{0,896}{22,4}=0,04\left(mol\right)=>m_{CO}=0,04.28=1,12\left(g\right)\)

a: \(n_{H_2}=\dfrac{V}{22.4}=1\left(mol\right)\)

b: \(n_{N_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)

Ta có: \(n_{CO_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

a. PTHH: CO₂ + 2NaOH ---> Na₂CO₃ + H₂O

b. Theo PT: \(n_{Na_2CO_3}=n_{CO_2}=0,5\left(mol\right)\)

=> \(m_{Na_2CO_3}=0,5.106=53\left(g\right)\)

c. (Không có bazơ sau phản ứng nên mik tìm m_NaOH nhé.)

Theo PT: \(n_{NaOH}=2.n_{CO_2}=2.0,5=1\left(mol\right)\)

=> \(m_{NaOH}=40.1=40\left(g\right)\)

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.

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