\(a,PTHH:2KClO_3\underrightarrow{t^o}2KCl+3O_2\\ Theo.PTHH:n_{O_2}=\dfrac{3}{2}.n_{KClO_2}=\dfrac{3}{2}.0,6=0,9\left(mol\right)\)
\(b,PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{H_2}=\dfrac{V_{\left(đktc\right)}}{22,4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ Theo.PTHH:n_{Mg}=n_{H_2}=0,1\left(mol\right)\\ m_{Mg}=n.M=0,1.24=2,4\left(g\right)\)
\(c,PTHH:CaCO_3\underrightarrow{t^o}CaO+CO_2\\ n_{CO_2}=\dfrac{V_{\left(đktc\right)}}{22,4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ Theo.PTHH:n_{CaCO_3}=n_{CO_2}=0,15\left(mol\right)\\ m_{CaCO_3}=n.M=0,15.100=15\left(g\right)\)
\(d,PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=\dfrac{m}{M}=\dfrac{2,8}{56}=0,05\left(mol\right)\\ Theo.PTHH:n_{H_2}=n_{Fe}=0,05\left(mol\right)\\ V_{H_2\left(đktc\right)}=n.22,4=0,05.22,4=1,12\left(l\right)\\ Theo.PTHH:n_{HCl}=2.n_{Fe}=2.0,05=0,1\left(mol\right)\\ m_{HCl}=n.M=0,1.36,5=3,65\left(g\right)\)
Câu 1:
a, 2KClO3 \(\underrightarrow{t^o}\) 2KCl + 3O2
\(n_{O_2}=\dfrac{0,6.3}{2}=0,9mol\)
b, Mg + 2HCl \(\rightarrow\) MgCl2 + H2
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\\ n_{Mg}=0,1mol\\ m_{Mg}=0,1.24=2,4g\)
c, CaCO3 \(\rightarrow\) CaO + CO2
\(n_{CO_2}=\dfrac{3,36}{22,4}=0,15mol\\ m_{CaCO_3}=0,15.100=15g\)
d, Fe + 2HCl \(\rightarrow\) FeCl2 + H2
\(n_{Fe}=\dfrac{2,8}{56}=0,05mol\)
\(V_{H_2}=0,05.22,4=1,12l\\ n_{HCl}=0,05.2=0,1mol\\ m_{HCl}=0,1.36,5=3,65g\)
