\(2KClO_3\xrightarrow[]{t^o}2KCl+3O_2\)

a) Ta có : \(n_{KCl}=\dfrac{14,9}{74,5}=0,2\left(mol\right)\)

Theo Pt : \(n_{KCl}=n_{KClO3}=0,2\left(mol\right)\Rightarrow m_{KClO3}=0,2.122,5=24,5\left(g\right)\)

b) Ta có : \(n_{KCl}=\dfrac{2,98}{74,5}=0,04\left(mol\right)\)

Theo Pt : \(n_{KCl}=n_{KClO3}=0,04\left(mol\right)\Rightarrow m_{KClO3}=0,04.122,5=4,9\left(g\right)\)

c) Theo Pt : \(n_{O2}=\dfrac{3}{2}n_{KClO3}=3,6\left(mol\right)\Rightarrow V_{O2\left(dkc\right)}=3,6.24,79=89,244\left(l\right)\)

\(n_{Ba\left(OH\right)_2}=0,1.1=0,1\left(mol\right)\)

\(n_{H_2SO_4}=0,1.0,8=0,08\left(mol\right)\)

PTHH:

\(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4+H_2O\)

0,08             0,08           0,08       0,08

\(\dfrac{0,1}{1}>\dfrac{0,08}{1}\)--> Ba(OH)2 dư 

\(m_{BaSO_4}=0,08.233=18,46\left(g\right)\)

\(Fe\left(OH\right)_2+2HCl\rightarrow FeCl_2+2H_2O\)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(n_{HCl}=\dfrac{25,55}{35,6}=0,7\left(mol\right)\)

PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,7}{3}\), ta được HCl dư.

Theo PT: \(n_{HCl\left(pư\right)}=3n_{Al}=0,6\left(mol\right)\Rightarrow n_{HCl\left(dư\right)}=0,7-0,6=0,1\left(mol\right)\)

\(\Rightarrow m_{HCl\left(dư\right)}=0,1.36,5=3,65\left(g\right)\)

\(Fe\left(OH\right)_2+H_2SO_4\rightarrow FeSO_4+2H_2O\)

\(a,n_{CuO}=\dfrac{m_{CuO}}{M_{CuO}}=\dfrac{8}{80}=0,1\left(mol\right)\\ b,n_{Fe_2\left(SO_4\right)_3}=\dfrac{m_{Fe_2\left(SO_4\right)_3}}{M_{Fe_2\left(SO_4\right)_3}}=\dfrac{20}{400}=0,05\left(mol\right)\)

\(a.n_{CuO}=\dfrac{8}{80}=0,1mol\\ b.n_{Fe_2\left(SO_4\right)_3}=\dfrac{20}{400}=0,05mol\)

\(CaCO_3\xrightarrow[t^0]{}CaO+CO_2\\ \Rightarrow\dfrac{1}{100}=\dfrac{m_{CaO\left(LT\right)}}{56}\\ \Rightarrow m_{CaO\left(LT\right)}=0,56\left(tấn\right)\\ m_{CaO\left(TT\right)}=0,56.90\%=0,504\left(tấn\right)\)

\(a.m_{H_2SO_4}=0,75.98=73,5g\)

\(b.m_{Cu}=0,6.64=38,4g\\ c.m_{\left(NH_4\right)_2SO_4}=0,8.132105,6g\)

\(n_{NaOH}=0,2.4=0,8\left(mol\right)\)

PT: \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)

\(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,4\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=0,4.98=39,2\left(g\right)\)

\(\Rightarrow C\%_{H_2SO_4}=\dfrac{39,2}{50}.100\%=78,4\%\)

a, \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\Rightarrow V_{H_2}=0,3.24,79=7,437\left(l\right)\)

\(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,1\left(mol\right)\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2\left(g\right)\)

b, \(n_{H_2SO_4}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\)

\(\Rightarrow V_{ddH_2SO_4}=\dfrac{0,3}{2}=0,15\left(l\right)\)