Bài 18: Mol

\(a.m_{Mg}=0,1.24=2,4\left(g\right)\\ m_{Ca}=0,2.40=8\left(g\right)\\ b.n_{hh}=\dfrac{2,8}{28}+\dfrac{13,2}{44}=0,4\left(mol\right)\\ \Rightarrow V_{hh}=0,4.22,4=8.96\left(l\right)\)

\(a,M_R=\dfrac{6}{0,15}=40\left(g/mol\right)\\ b,M_A=\dfrac{m_A}{n_A}=\dfrac{7}{\dfrac{5,6}{22,4}}=\dfrac{7}{0,25}=28\left(g/mol\right)\\ c,\overline{M_{hh}}=\dfrac{4\cdot28+1\cdot32}{4+1}=\dfrac{144}{5}=28,8\left(g/mol\right)\)

a) \(n_{Fe_3O_4}=\dfrac{4,64}{232}=0,02\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Fe}=3n_{Fe_3O_4}=0,06\left(mol\right)\\n_O=4n_{Fe_3O_4}=0,08\left(mol\right)\end{matrix}\right.\)

b) \(n_{N_2O}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_N=2n_{N_2O}=0,3\left(mol\right)\\n_O=n_{N_2O}=0,15\left(mol\right)\end{matrix}\right.\)

c) Ta có: \(n_{H_2SO_4}=\dfrac{4,9}{98}=0,05\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_H=2n_{H_2SO_4}=0,1\left(mol\right)\\n_S=n_{H_2SO_4}=0,05\left(mol\right)\\n_O=4n_{H_2SO_4}=0,2\left(mol\right)\end{matrix}\right.\)

\(a,n_{CaCO_3}=\dfrac{50}{100}=0,5\left(mol\right)\\ PTHH:CaCO_3\rightarrow^{t^o}CaO+CO_2\\ \Rightarrow n_{CaO}=n_{CaCO_3}=0,5\left(mol\right)\\ b,n_{CO_2}=n_{CaCO_3}=0,5\left(mol\right)\\ \Rightarrow V_{CO_2\left(đktc\right)}=0,5\cdot22,4=11,2\left(l\right)\)

\(1,M_{CO_2}=12+16\cdot2=44\left(g/mol\right)\\ 2,M_{H_2SO_4}=2+32+16\cdot4=98\left(g/mol\right)\\ 3,M_{Cu\left(NO_3\right)_2}=64+\left(14+16\cdot3\right)\cdot2=188\left(g/mol\right)\\ 4,M_{Fe_2\left(SO_4\right)_3}=56\cdot2+\left(32+16\cdot4\right)\cdot3=400\left(g/mol\right)\\ 5,M_{Na_3PO_4}=23\cdot3+31+16\cdot4=164\left(g/mol\right)\)

a) Số nguyên tử Al trong 1 mol nguyên tử Al: 6.10²³ (nguyên tử)

b) Số phân tử O₂ có chứa 1 phân tử O₂ là: 6.10²³ (phân tử)

c) Số phân tử CO₂ trong 0,5 mol phân tử khí CO₂ : 6.10²³.0,5=3.10²³ (phân tử)

Ta có: \(n_{CO_2}=\dfrac{3,36}{22.4}=0,15\left(mol\right)\)

\(\Rightarrow m_{CO_2}=0,15.44=6,6\left(g\right)\)

nCO₂=3,36/22,4=0,15 mol

mCO₂=0,15.44=6,6 g

\(a,m_{Al\left(NO_3\right)_3}=1,5\cdot213=319,5\left(g\right)\\ b,n_{C_3H_8}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ \Rightarrow m_{C_3H_8}=0,3\cdot44=13,2\left(g\right)\)

Ta có: \(n_{CO_2}=\dfrac{11}{44}=0,25\left(mol\right)\)

\(\Rightarrow V_{CO_2}=0,25.22,4=5,6\left(l\right)\)

Bạn tham khảo nhé!

Ta có: \(n_{CO_2}=\dfrac{11}{44}=0,25\left(mol\right)\)

\(\Rightarrow V_{CO_2}=0,25.22,4=5,6\left(lít\right)\)

Ta có: \(n_{H_2}=\dfrac{6,1975}{22,4}\approx0,2767\left(mol\right)\)

Bạn tham khảo nhé!

\(n_{H_2}=\dfrac{6,1975}{22,4}\approx0,3\left(mol\right)\)