\(1.a.CaCO_3.t^o\rightarrow CaO+CO_2\\ b.m_{CaCO_3}=m_{CaO}+m_{CO_2}\\ \Rightarrow m_{CO_2}=m_{CaCO_3}-m_{CaO}=20-11,2=8,8\left(g\right)\)

n_CaCO3 = \(\dfrac{15}{100}=0,15\left(mol\right)\)

PTHH: CaCO3 -> CaO + CO2

PT:          1              1            1   (mol)

ĐB:          0,15       0,15       0,15 (mol)

m_CaO = 0,15.56 = 8,4 (g)

V_CO2  = 0,15.22,4 = 3,36(l)

ảo thặc 100 caco3 lun:>

PTPƯ:

\(CaCO_3\underrightarrow{t^o}CaO+CO_2\)\(\uparrow\)

0,015      0,015      0,015

\(nCaCO_3=\dfrac{1,5}{100}=0,015mol\)

\(mCaO=0,015.56=0,84\)(tấn)

a, CaCO3 --t0--> CaO + CO2

b, Ta có nCO2= 5,6/22,4=0,25 = nCaCO3

=> mCaCO3= 0,25.100=25 g

c, nCaO = nCO2=0,25

=> mCaO= 0,25.56=14 g

a, PTHH : CaCO3 \(\underrightarrow{t^o}\) CaO+CO2

b,n CaCO3= n CO2 =\(\frac{5,6}{22,4}\)= 0,25 g

=>m CaCO3 = 0,25.100 = 0,5g

c, m CaO= 0,25. 56 = 14 ( g)

\(PTHH:CaCO_3\rightarrow CaO+CO_2\)

- 100g/mol 56g/mol

- 50g ?

\(m_{CaO}=\frac{59.56}{100}=28\left(g\right)\)

\(m_{CaO_{tt}}=\frac{28.90\%}{100\%}=25,5\left(g\right)\)

a)CaCO₃\(\xrightarrow[]{t\text{°}}\)CaO+CO₂.

b)Áp dụng định luật bảo toàn khối lượng:

\(m_{CaCO_3}=m_{CaO}+m_{CO_2}\)

\(m_{CaCO_3}=5,6+4,4\)

\(m_{CaCO_3}=10\left(\text{tấn}\right)\)

a) PTHH: \(CaCO_3\underrightarrow{t^o}CaO+CO_2\)

b) m_CaCO3 = \(\frac{120.90}{100}=108\left(gam\right)\)

=> n_CaCO3 = 108 / 100 = 1,08 (mol)

=> n_CaO = n_CO2 = n_CaCO3 = 1,08 (mol)

=> m_CaO = 1,08 x 56 = 60,28 (gam) = a

=> V_CO2(đktc) = 1,08 x 22,4 = 24,192 (lít) = b

\(CaCO_3\xrightarrow[]{t^0}CaO+CO_2\)

Cứ 100 tấn đá vôi thì nung được 56 tấn vôi

\(\Rightarrow\dfrac{1}{100}=\dfrac{m_{CaO\left(lt\right)}}{56}\\ \Rightarrow m_{CaO\left(lt\right)}=0,56tấn\\ m_{CaO\left(tt\right)}=0,56.70\%=0,392tấn\)

\(n_{MgCO_3}=a\left(mol\right)\)

\(n_{CaCO_3}=b\left(mol\right)\)

\(n_{CO_2}=\dfrac{11.2}{22.4}=0.5\left(mol\right)\)

\(MgCO_3\underrightarrow{^{^{t^0}}}MgO+CO_2\)

\(CaCO_3\underrightarrow{^{^{t^0}}}CaO+CO_2\)

\(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)

\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

\(\left\{{}\begin{matrix}a+b=0.5\\40a+56b=2.2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=1.6125\\b=-1.1125\end{matrix}\right.\)

Xem lại đề !

\(Đặt:m_{Al_2O_3}=g\left(g\right)\Rightarrow m_{CaCO_3}+m_{MgCO_3}=8g\left(g\right)\\ PTHH:CaCO_3\rightarrow\left(t^o\right)CaO+CO_2\\ MgCO_3\rightarrow\left(t^o\right)CO_2+MgO\\ Đặt:n_{CaCO_3}=a\left(mol\right);n_{MgCO_3}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}100a+84b=8g\\56a+40b+g=60\%.9g=5,4g\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{31}{440}g\\b=\dfrac{1}{88}g\end{matrix}\right.\\ \Rightarrow\%m_{Al_2O_3}=\dfrac{g}{9g}.100\%=11,111\%\)

\(\%m_{MgCO_3}=\dfrac{\dfrac{1}{88}g.84}{9g}.100\%\approx10,606\%\\ \%m_{CaCO_3}=\dfrac{\dfrac{31}{440}g.100}{9g}.100\approx78,283\%\)

\(b,\%m_{\dfrac{Al_2O_3}{A}}=\dfrac{g}{\dfrac{1}{88}.40g+\dfrac{31}{440}.56g+g}.100\approx18,5185\%\\ \%m_{\dfrac{MgO}{A}}=\dfrac{\dfrac{1}{88}.40g}{\dfrac{1}{88}.40g+\dfrac{31}{440}.56g+g}.100\approx8,4175\%\\ \Rightarrow m_{Al_2O_3}=18,5185\%.2=0,37037\left(g\right)\Rightarrow n_{Al_2O_3}=\dfrac{0,37037}{102}\left(mol\right)\\ m_{MgO}=8,4175\%.2=0,16835\left(g\right)\\ \Rightarrow n_{MgO}=\dfrac{0,16835}{40}\left(mol\right)\\ m_{CaO}=2-\left(0,37037+0,16835\right)=1,46128\left(g\right)\\ \Rightarrow n_{CaO}=\dfrac{1,46128}{56}\left(mol\right)\)

\(PTHH:Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\\ CaO+2HCl\rightarrow CaCl_2+H_2O\\ MgO+2HCl\rightarrow MgCl_2+H_2O\\ n_{HCl}=2.\left(\dfrac{1,46128}{56}+\dfrac{0,16835}{40}\right)+6.\dfrac{0,37037}{102}\approx0,0824\left(mol\right)\\ \Rightarrow V_{ddHCl}\approx\dfrac{0,0824}{0,5}\approx0,1648\left(lít\right)\approx164,8\left(ml\right)\)