\(a,m_{NaOH}=0,25.40=10(g)\\ b,n_{CO_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ m_{CO_2}=0,1.44=4,4(g)\)

m= n.M 

=> mO₂ = 0,15.32= 4,8 gam

mNaOH = 0,4.40 = 16 gam

nCO₂ = \(\dfrac{1,68}{22,4}\)= 0,075 mol => mCO₂ = 0,075.44 = 3,3 gam

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.

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

a)n CO=0,1 mol

n CO2=0,15 mol

=>Dhh\MO2=(0,1.28+0,15.44)\32=0,293

n NO2=0,25 =>m=0,25.46=11,5g

n CO2=0,4 mol=>m=0,4.44=17,6g

=>Dhh\MSO2=0,4546

a. \(m_{CO_2}=\dfrac{8,96}{22,4}.44=17,6\left(g\right)\)

b. \(m_{Fe}=\dfrac{1,8.10^{22}}{6.10^{23}}.56=1,68\left(g\right)\)

c. \(m_{Fe_2\left(SO_4\right)_3}=0,25.400=100\left(g\right)\)

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

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

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) mO2= nO2. M(O2)=0,45. 32=14,4(g)

b) mBaCO3=nBaCO3.M(BaCO3)=0,6.197=118,2(g)

c) mAl2(SO4)3=nAl2(SO4)3.M(Al2(SO4)3)=1,5.342=513(g)

d) nSO2=V(SO2,đktc)/22,4=16,8/22,4=0,7(mol)

=> mSO2=nSO2.M(SO2)=0,7.64=44,8(g)

e) nH2O=(3.10²³):(6.10²³)=0,5(mol)

=>mH2O=nH2O.M(H2O)=0,5.18=9(g)

f) nCO2=V(CO2,đktc)/22,4=8,96/22,4=0,4(mol)

=>mCO2=nCO2.M(CO2)=0,4.44=17,6(g)

\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25mol\)

\(\Rightarrow n_{O_2}=0,25mol\)         \(\Rightarrow m_{O_2}=0,25.32=8g\)