a, m_CaO = 0,5.56 = 28 (g)

b, \(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

c, \(n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\left(mol\right)\)

d, \(V_{hhk}=0,2.22,4+0,3.22,4=11,2\left(l\right)\)

e, \(\%m_{Cu}=\dfrac{64}{64+32+16.4}.100\%=40\%\)

Bạn tham khảo nhé!

a) mCaO=nCaO.M(CaO)=0,5.56=28(g)

b) nCO2=V(CO2,dktc)=6,72/22.4=0,3(mol)

c) nH2SO4=mH2SO4/M(H2SO4)=24,5/98=0,25(mol)

d) V(hh H2,NH3)=(0,3+0,2).22,4=11,2(l)

e) %mCu/CuSO4=(64/160).100=40%

Chúc em học tốt!

a) \(n_{N_2}+n_{CO_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\)

Có: \(28.n_{N_2}+44.n_{CO_2}=24,4\)

=> \(\left\{{}\begin{matrix}n_{N_2}=0,4\left(mol\right)\\n_{CO_2}=0,3\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%V_{N_2}=\dfrac{0,4}{0,7}.100\%=57,143\%\\\%V_{CO_2}=\dfrac{0,3}{0,7}.100\%=42,857\%\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}m_{N_2}=0,4.28=11,2\left(g\right)\\m_{CO_2}=0,3.44=13,2\left(g\right)\end{matrix}\right.\)

a) Gọi số mol N₂, O₂ trong 6,72l khí A lần lượt là a, b

=> \(\left\{{}\begin{matrix}28a+32b=8,8\\a+b=\dfrac{6,72}{22,4}=0,3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,1\left(mol\right)\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%V_{N_2}=\dfrac{0,2}{0,3}.100\%=66,67\%\\\%V_{O_2}=\dfrac{0,1}{0,3}.100\%=33,33\%\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{N_2}=\dfrac{28.0,2}{8,8}.100\%=63,64\%\\\%m_{O_2}=\dfrac{32.0,1}{8,8}.100\%=36,36\%\end{matrix}\right.\)

b) 

\(n_A=0,3\left(mol\right)\)

\(\Rightarrow n_{CO_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{CO_2}=0,3.44=13,2\left(g\right)\)

c) 2,2g A có thể tích là 1,68 lít

=> \(V_{H_2}=1,68\left(l\right)\)

a, \(n_{C_2H_4}+n_{C_2H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\left(1\right)\)

PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

Theo PT: \(n_{Br_2}=n_{C_2H_4}+2n_{C_2H_2}=\dfrac{72}{160}=0,45\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{C_2H_4}=0,15\left(mol\right)\\n_{C_2H_2}=0,15\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{C_2H_4}=\dfrac{0,15.28}{0,15.28+0,15.26}.100\%\approx51,85\%\\\%m_{C_2H_2}\approx48,15\%\end{matrix}\right.\)

\(\%V_{C_2H_4}=\%V_{C_2H_2}=\dfrac{0,15.22,4}{6,72}.100\%=50\%\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{C_2H_4Br_2}=n_{C_2H_4}=0,15\left(mol\right)\\n_{C_2H_2Br_4}=n_{C_2H_2}=0,15\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{C_2H_4Br_2}=0,15.188=28,2\left(g\right)\\m_{C_2H_2Br_4}=0,15.346=51,9\left(g\right)\end{matrix}\right.\)

Đáp án : A

+) HCl : Al -> 1,5H₂

               Mg -> H₂

+) NaOH : Al -> 1,5H₂

=> n_Al = 2/3 n_H2 = 0,2 mol

=> n_Mg = n_H2(1) – n_H2(Al) = 0,4 – 0,3 = 0,1 mol

=> %m_Al = 69,23%

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

PTHH : \(C+O_2\rightarrow CO_2\)

            x        x        x        (mol)

PTHH : \(S+O_2\rightarrow SO_2\)

           y          y          y       (mol)

\(\rightarrow\) x + y = 0,3 (1)

12x+32y=5,6  (2)

Từ (1),(2) \(\rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)

\(\%C=\dfrac{0,2.12}{5,6}.100=42,86\%\)

\(\%S=100\%-42,86\%=57,14\%\)

\(m_C=0,2.12=2,4\left(g\right)\)

\(m_S=5,6-2,4=3,2\left(g\right)\)