a, \(CaO+H_2SO_4\rightarrow CaSO_4+H_2O\)

b, \(n_{CaO}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{CaSO_4}=n_{CaO}=0,1\left(mol\right)\)

\(\Rightarrow m_{CaSO_4}=0,1.136=13,6\left(g\right)\)

c, \(m_{H_2SO_4}=0,1.98=9,8\left(g\right)\)

a) \(n_{CaO}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

\(CaO+H_2SO_4\)\(\rightarrow CaSO_4+H_2O\)

b) Ta có \(n_{CaSO_4}=n_{CaO}=0,1\left(mol\right)\)

\(\rightarrow m_{CaSO_4}=0,1\cdot136=13,6\left(g\right)\)

c) Ta có \(n_{H_2SO_4}=n_{CaO}=0,1\left(mol\right)\)

\(\rightarrow m_{H_2SO_4}=0,1\cdot98=9,8\left(g\right)\)

Zn+2HCl-->ZnCl2+H2

a, nZn=\(\dfrac{19,5}{65}\)=0,3mol

nH2=nZn=0,3mol

VH2=0,3*22,4=6,72l

b, nZnCl2=nZn=0,3mol

mZnCl2=0,3*(65+35,5*2)=40,8g

c,nHCl=2nZn=0,6mol

mHCl=0,6*36,5=21,9g

C%HCl=\(\dfrac{21.9}{200}\)*100%=10,95%

a) pthh: Zn + 2Hcl = \(ZnCl_2\) + \(H_2\)

\(_{_{ }}\)\(N_{ZN}\) = \(\dfrac{m}{M}\) =\(\dfrac{19,5}{65}\) =0.3 mol 

\(N_{H_2}\)= \(N_{Zn}\) = 0.3 mol 

\(V_{H_2}\)= n × 24.79 = 0.3 × 24.79 = 7.437 ( L)

b) \(N_{ZnCl_2}\)= \(N_{Zn}\) = 0.3 mol 

\(m_{ZnCl_2}\)= n × M = 0.3 × 136 = 40.8 g

c) \(m_{dd}\) = 19.5 + 200 = 219.5 g

\(C\%\:=\dfrac{m_{Ct}}{m_{dd}}\) × 100 =\(\dfrac{19.5}{219.5}\)×100= 8.88 %

a, \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

Ta có: \(n_{C_2H_4Br_2}=\dfrac{18,8}{188}=0,1\left(mol\right)\)

Theo PT: \(n_{C_2H_4}=n_{Br_2}=n_{C_2H_4Br_2}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,1.22,4}{4,48}.100\%=50\%\\\%V_{CH_4}=50\%\end{matrix}\right.\)

b, \(C\%_{Br_2}=\dfrac{0,1.160}{100}.100\%=16\%\)

Gọi CTPT của alcohol là \(C_nH_{2n+2}O\left(n\ge1\right)\)

PTHH: \(2C_nH_{2n+2}O+2Na\rightarrow2C_nH_{2n+1}ONa+H_2\)

Ta có \(n_{H_2}=\dfrac{V_{H_2}}{24,79}=\dfrac{3,09875}{24,79}=0,125\left(mol\right)\)

\(\Rightarrow n_{alcohol}=0,25\left(mol\right)\)

\(\Rightarrow M_{alcohol}=\dfrac{m}{n}=\dfrac{15}{0,25}=60\left(g\right)\)

Mà \(M_{alcohol}=14n+18\)

\(\Rightarrow14n+18=60\) \(\Leftrightarrow n=3\)

Vậy CTPT của alcohol là \(C_3H_8O\)