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.

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

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

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

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

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

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

\(V_{H_2}=n\cdot22,4=2\cdot22,4=44,8\left(l\right)\\ =>A\)

mMg = 0,5.24 = 12 gam

V_SO2 = n.22,4 = 0,25.22,4 = 5,6 lít

nN₂ = \(\dfrac{16,8}{22,4}\)= 0,75 mol , nO₂ = \(\dfrac{5,6}{22,4}\)= 0,25 mol

=> m(N₂  + O₂ ) = 0,75.28 + 0,25.32 = 29 gam

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

1. B

2. B

(Câu 2 cậu nên sửa lại câu hỏi nhé: Khối lượng dung dịch NaOH 10% ...)

Câu 1.

\(n_{Fe}=\dfrac{5,6}{56}=0,1mol\)

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

0,1                                 0,1

\(V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\)

Chọn B.

Câu 2.  \(n_{HCl}=0,2\cdot1=0,2mol\)  

       Để trung hòa: \(\Rightarrow n_{H^+}=n_{OH^-}=0,2\)

       \(m_{NaOH}=0,2\cdot40=8\left(g\right)\)

       \(m_{ddNaOH}=\dfrac{8}{10\%}\cdot100\%=80\left(g\right)\)

     Chọn B.

C

A

c

Fe+2HCl->FeCl₂+H₂

0,1--0,2------0,1----0,1 mol

nFe=\(\dfrac{5,6}{56}=0,1mol\)

m ddHCl=50g

=>m dd tạo thành là : 5,6+50-0,1.2=55,4g

c) Zn+H2SO4->ZnSO4+H2

    1-----------------------------1 mol

n H2=1 mol

=>mZn=1.65=65g

#Check lại ý b

CTVHOC24

\(a.\)

\(b.\)

\(\Rightarrow m_{SO2}=0,25\times64=16\left(gam\right)\)

\(\Rightarrow m_{H2}=1\times2=2\left(gam\right)\)

a) \(n_{Fe}=\frac{m}{M}=\frac{11,2}{56}=0,2\left(mol\right)\)

\(n_{H_2SO_4}=\frac{m}{M}=\frac{19,6}{98}=0,2\left(mol\right)\)

b) \(n_{SO_2}=\frac{V}{22,4}=\frac{5,6}{22,4}=0,25\left(mol\right)\)

\(\Rightarrow m_{SO_2}=M.n=64.0,25=16\left(g\right)\)

* \(n_{H_2}=\frac{V}{22,4}=\frac{22,4}{22,4}=1\left(mol\right)\)

\(\Rightarrow m_{H_{ }_2}=M.n=2.1=2\left(g\right)\)