a) \(n_{Cu}=\dfrac{12,8}{64}=0,2\left(mol\right)\)

b) \(n_{CO_2}=\dfrac{66}{44}=1,5\left(mol\right)\)

c) \(n_{Fe_2\left(SO_4\right)_3}=\dfrac{25}{400}=0,0625\left(mol\right)\)

d) \(n_{Fe_2O_3}=\dfrac{8}{160}=0,05\left(mol\right)\)

e) \(n_{NH_3}=\dfrac{1,8.10^{22}}{6.10^{23}}=0,03\left(mol\right)\)

f) \(n_{Cl_2}=\dfrac{20,16}{22,4}=0,9\left(mol\right)\)

Đề yêu cầu tính gì á em?

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a. \(n_{Ag}=\dfrac{1,8.10^{25}}{6.10^{23}}=30\left(mol\right)\)

b. \(n_{CO_2}=\dfrac{59,4}{44}=1,35\left(mol\right)\)

c. \(n_{K_2O}=\dfrac{4,2.10^{22}}{6.10^{23}}=0,07\left(mol\right)\)

d. \(n_{CuSO_4}=\dfrac{18.10^{23}}{6.10^{23}}=3\left(mol\right)\)

e. \(n_{SO_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)

g. \(n_{Fe_3O_4}=\dfrac{52,2}{232}=0,225\left(mol\right)\)

h. \(n_{O_2}=\dfrac{6,72}{22,4}-0,3\left(mol\right)\)

i. \(n_{N_2}=\dfrac{13,6}{22,4}\approx0,6\left(mol\right)\)

$n_{CO_2} = \dfrac{8,8}{44} = 0,2(mol)$
$V_{CO_2} = 0,2.22,4 = 4,48(lít)$

$n_{NO_2} = \dfrac{6.10^{23}}{6.10^{23}} = 1(mol)$
$V_{NO_2} = 1.22,4 = 22,4(lít)$

$n_{SO_2} = \dfrac{12,8}{64} = 0,2(mol)$
$V_{SO_2} = 0,2.22,4 = 4,48(lít)$
$n_{SO_3} = \dfrac{1,5.10^{23}}{6.10^{23}} = 0,25(mol)$
$V_{SO_3} = 0,25.22,4 = 5,6(lít)$

Hãy tính thể tích (đktc) của

b)8,8g CO2

\(n_{CO_2}=\dfrac{8,8}{44}=0,2\left(mol\right)\Rightarrow V_{CO_2}=0,2.22,4=4,48\left(l\right)\)

C)6.10²³ phân tử NO2

\(n_{NO_2}=\dfrac{6.10^{23}}{6.10^{23}}=1\Rightarrow V_{NO_2}=1.22,4=22,4\left(lít\right)\)

d)Hỗn hợp gồm {12,8 g SO2, 1,5.10 pt SO3

\(n_{SO_2}=\dfrac{12,8}{54}=0,2\left(mol\right);n_{SO3}=\dfrac{1,5.10^{23}}{6.10^{23}}=0,25\left(mol\right)\)

=> \(V_{hh}=\left(0,2+0,25\right).22,4=10,08\left(l\right)\)

a) 

\(n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)\)

\(n_{Ca}=\dfrac{20}{40}=0,5\left(mol\right)\)

\(n_{CaCO_3}=\dfrac{25}{100}=0,25\left(mol\right)\)

\(n_{NaOH}=\dfrac{4}{40}=0,1\left(mol\right)\)

\(n_{H_2O}=\dfrac{1,5.10^{23}}{6.10^{23}}=0,25\left(mol\right)\)

b) 

\(m_{ZnSO_4}=0,25.161=40,25\left(g\right)\)

\(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)

\(m_{Cu}=0,3.64=19,2\left(g\right)\)

\(m_{Fe_2\left(SO_4\right)_3}=0,35.400=140\left(g\right)\)

c)

\(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)

\(V_{Cl_2}=0,15.22,4=3,36\left(l\right)\)

\(V_{SO_2}=0,3.22,4=6,72\left(l\right)\)

\(V_{CH_4}=0,5.22,4=11,2\left(l\right)\)

a: \(n_{Fe}=\dfrac{14}{56}=0.25\left(mol\right)\)

\(n_{Ca}=\dfrac{20}{40}=0.5\left(mol\right)\)

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)

Chọn A