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\(PTHH:MgCO_3\rightarrow MgO+CO_2\)
Ta có:
\(\%m_{MgO}=\frac{m}{m_{dd}}.100=\frac{4}{4+4,4}.100=47,62\%\)
\(\%m_{CO_2}=100-47,62=52,38\%\)
Vậy .............
\(a,n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ PTHH:2K+2H_2O\rightarrow2KOH+H_2\uparrow\\ Theo.pt:n_K=2n_{H_2}=2.0,1=0,2\left(mol\right)\\ m_K=0,2.39=7,8\left(g\right)\\ m_{K_2O}=17,2-7,8=9,4\left(g\right)\\ b,n_{CuO\left(bđ\right)}=\dfrac{12}{80}=0,15\left(mol\right)\\ PTHH:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ LTL:0,15>0,1\Rightarrow Cu.dư\)
Gọi nCuO (pư) = a (mol)
=> nCu = a (mol)
mchất rắn sau pư = 80(0,15 - a) + 64a = 10,8
=> a = 0,075 (mol)
=> nH2 (pư) = 0,075 (mol)
\(H=\dfrac{0,075}{0,1}=75\%\)
\(a.n_{CuO}=\dfrac{40}{80}=0,5\left(mol\right)\\ n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ PTHH:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ Vì:\dfrac{0,15}{1}< \dfrac{0,5}{1}\\ \rightarrow CuOdư\\ n_{CuO\left(p.ứ\right)}=n_{Cu}=n_{H_2}=0,15\left(mol\right)\\ \rightarrow n_{CuO\left(dư\right)}=0,5-0,15=0,35\left(mol\right)\\ m_{CuO\left(DƯ\right)}=0,35.80=28\left(g\right)\\ b.m_{Cu}=0,35.64=22,4\left(g\right)\\ c.m_{hh_{rắn}}=m_{Cu}+m_{CuO\left(dư\right)}=22,4+28=50,4\left(g\right)\)
Gọi \(n_{CaCO_3}=a\left(mol\right)\) và \(n_{MaCO_3}=b\left(mol\right)\)
PTHH: \(CaCO_3\underrightarrow{t^o}CaO+CO_2\)
\(MgCO_3\underrightarrow{t^o}MgO+CO_2\)
\(\Rightarrow m_{hh}=100a+84b=18,4\)
\(n_{CO_2}=\dfrac{8,8}{44}=0,2\left(mol\right)\Rightarrow a+b=0,2\left(mol\right)\)
\(\Rightarrow a=b=0,1\left(mol\right)\)
\(\Rightarrow m_{CaCO_3}=10g;m_{MgCO_3}=8,4g\)
\(\Rightarrow\%m_{CaCO_3}=\dfrac{100\%.10}{18,4}\approx54\%;\%m_{MgCO_3}=100\%-54\%=46\%\)
a) PTHH: \(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\uparrow\)
b) Ta có: \(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)=n_{NaOH}\) \(\Rightarrow m_{NaOH}=0,1\cdot40=4\left(g\right)\)
c) PTHH: \(H_2+CuO\xrightarrow[]{t^o}Cu+H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{H_2}=0,05\left(mol\right)\\n_{CuO}=\dfrac{10}{80}=0,125\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) CuO còn dư, Hidro p/ứ hết
\(\Rightarrow\left\{{}\begin{matrix}n_{Cu}=0,05\left(mol\right)\\n_{CuO\left(dư\right)}=0,075\left(mol\right)\end{matrix}\right.\) \(\Rightarrow m_{rắn}=m_{Cu}+m_{CuO}=9,2\left(g\right)\)
\(Fe_2O_3 + 3CO \xrightarrow{t^o} 2Fe + 3CO_2\\ CuO + CO \xrightarrow{t^o} Cu + CO_2\\ CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O\\ n_{CO} = n_{CO_2} = n_{CaCO_3} = \dfrac{15}{100} = 0,15(mol)\\ \Rightarrow m = m_Y + m_{CO_2} - m_{CO} = 200 + 0,15.44 - 0,15.28 = 202,4(gam)\)
goi a la khoi luong cua CaCO3
b la khoi luong cua MgCO3
\(n_{CaCO_3}=\dfrac{a}{100}\left(mol\right)\)
\(n_{MgCO_3}=\dfrac{b}{84}\left(mol\right)\)
\(CaCO_3\rightarrow CaO+CO_2\) (1)
de: \(\dfrac{a}{100}\rightarrow\dfrac{a}{100}\rightarrow\dfrac{a}{100}\left(mol\right)\)
\(MgCO_3\rightarrow MgO+CO_2\) (2)
de: \(\dfrac{b}{84}\rightarrow\dfrac{b}{84}\rightarrow\dfrac{b}{84}\left(mol\right)\)
theo de: \(m_{CaO}+m_{MgO}=\dfrac{a+b}{2}\)
\(\Rightarrow0,56a+\dfrac{10}{21}b=\dfrac{a+b}{2}\) (3)
theo ĐLBTKL: \(m_{CO_2}=m_{CO_2\left(1\right)}+m_{CO_2\left(2\right)}=\dfrac{a+b}{2}\)
\(\Leftrightarrow0,44a+\dfrac{11}{21}b=\dfrac{a+b}{2}\) (4)
tu (3) va (4) \(\Rightarrow0,56a+\dfrac{10}{21}b=0,44a+\dfrac{11}{21}b\)
\(\Leftrightarrow0,12a=\dfrac{1}{21}b\)
\(\Leftrightarrow\dfrac{a}{b}=\dfrac{1}{21}:0,12=\dfrac{25}{63}\)
\(\Rightarrow m_{CaCO_3}=25\) g va \(m_{MgCO_3}=63\) g
\(\%m_{CaCO_3}=\dfrac{25}{25+63}.100\approx28,4\%\)
\(\%m_{MgCO_3}=100\%-\%m_{CaCO_3}=100-28,4=71,6\%\)
minh lm hoi tat nen ban thong cam
