\(n_{CaCO_3}=\dfrac{50}{100}=0,5\left(mol\right)\\ a,PTHH:CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ b,n_{CO_2}=n_{CaCl_2}=n_{CaCO_3}=0,5\left(mol\right)\\ V_{CO_2\left(25^oC,1bar\right)}=0,5.24,79=12,395\left(l\right)\\ c,n_{HCl}=2.0,5=1\left(mol\right)\\ V_{ddHCl}=\dfrac{1}{0,2}=5\left(l\right)\\ d,m_{CaCl_2}=111.0,5=55,5\left(g\right)\)

\(n_{Na_2CO_3.10H_2O}=\dfrac{85,8}{286}=0,3mol\\ m_{Na}=0,3.46=13,8g\\ m_C=0,3.12=3,6g\\ m_O=0,3.208=62,4g\\ m_H=0,3.20=6g\)

Ta có: \(n_{KOH}=0,02.0,2=0,004\left(mol\right)\)

PT: \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)

______0,004_____0,002____0,002 (mol)

\(\Rightarrow a=m_{K_2SO_4}=0,002.174=0,348\left(g\right)\)

\(V=V_{H_2SO_4}=\dfrac{0,002}{0,5}=0,004\left(l\right)=4\left(ml\right)\)

\(C_nH_{2n-2}+\dfrac{3n-1}{2}O_2\underrightarrow{t^o}nCO_2+\left(n-1\right)H_2O\)

Câu 1
\(a)PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ b)200ml=0,2l\\ n_{HCl}=0,2.1=0,2mol\\ n_{H_2}=n_{MgCl_2}=\dfrac{1}{2}n_{H_2}=\dfrac{1}{2}\cdot0,2=0,1mol\\ V_{H_2}=0,1.24,79=2,479l\\ c)C_{M_{MgCl_2}}=\dfrac{0,1}{0,2}=0,5M\)

Câu 2
\(a)PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ b)n_{Mg}=\dfrac{1,2}{24}=0,05mol\\ n_{H_2}=n_{MgCl_2}=n_{Mg}=0,05mol\\ V_{H_2}=0,05.24,79=1,2395l\\ c)n_{HCl}=2n_{Mg}=2.0,05=0,1mol\\ V_{ddHCl}=\dfrac{0,1}{2}=0,05l\\ d)C_{M_{MgCl_2}}=\dfrac{0,05}{0,05}=1M\)

\(n_{BaCl_2}=\dfrac{200.20,8\%}{208}=0,2\left(mol\right)\\ PTHH:BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\\ n_{BaSO_4}=n_{H_2SO_4}=n_{BaCl_2}=0,2\left(mol\right)\\ a,m_{kt}=m_{BaSO_4}=233.0,2=46,6\left(g\right)\\ b,C\%_{ddH_2SO_4}=\dfrac{0,2.98}{200}.100\%=9,8\%\)

\(200ml=0,2l\\ n_{Na_2SO_4}=0,2.0,1=0,02mol\\ a)Na_2SO_4+BaCl_2\rightarrow BaSO_4+2NaCl\)
0,02                 0,02          0,02             0,04
\(b)m=m_{BaSO_4}=0,02.233=4,66g\\ c)50ml=0,05l\\ C_{M_{BaCl_2}}=\dfrac{0,02}{0,05}=0,4M\)

\(\left(1\right)C+O_2\xrightarrow[]{t^0}CO_2\\ \left(3\right)CO_2+Na_2O\rightarrow Na_2CO_3\\ \left(3\right)Na_2CO_3+CaCl_2\rightarrow CaCO_3+2NaCl\)

\(a.Fe+2HCl\rightarrow FeCl_2+H_2\\ b.H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)

Ta có: \(n_{Fe}=\dfrac{6,72}{56}=0,12\left(mol\right)\)

PT: \(2Fe+O_2\underrightarrow{t^o}2FeO\)

Theo PT: \(\left\{{}\begin{matrix}n_{FeO}=n_{Fe}=0,12\left(mol\right)\\n_{O_2}=\dfrac{1}{2}n_{Fe}=0,06\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow m_{FeO}=0,12.72=8,64\left(g\right)\)

\(V_{O_2}=0,06.24,79=1,4874\left(l\right)\)