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\(m_{CaO\left(lt\right)}=\dfrac{94,08}{80\%}\cdot100\%=117,6kg\\ CaCO_3\xrightarrow[]{t^0}CaO+CO_2\\ \Rightarrow\dfrac{m_{CaCO_3}}{100}=\dfrac{117,6}{56}\\ \Rightarrow m_{CaCO_3}=210kg\\ \%m_{CaCO_3\left(trong.đá.vôi\right)}=\dfrac{210}{280}\cdot100\%=75\%\)
PT: \(CaCO_3\underrightarrow{t^o}CaO+CO_2\)
Ta có: \(n_{CaO}=\dfrac{94,08}{56}=1,68\left(kmol\right)\)
Theo PT: \(n_{CaCO_3\left(LT\right)}=n_{CaO}=1,68\left(kmol\right)\)
Mà: H = 80%
\(\Rightarrow n_{CaCO_3\left(TT\right)}=\dfrac{1,68}{80\%}=2,1\left(mol\right)\)
\(\Rightarrow m_{CaCO_3\left(TT\right)}=2,1.100=210\left(kg\right)\)
\(\Rightarrow\%CaCO_3=\dfrac{210}{280}.100\%=75\%\)
1)
1,2 tấn = 1200(kg)
5 tạ = 500(kg)
mCaCO3=1200.80%=960(kg)mCaCO3=1200.80%=960(kg)
CaCO3to→CaO+CO2nCaCO3 pư=nCaO=50056(mol)⇒H=50056.100960.100%=93%
\(m_{\text{CaCO_3}}=1000.95\%=950kg\\ \rightarrow n_{\text{CaCO_3}}=9,5mol\)
\(m_{CaCO_3}\underrightarrow{t^o}CaO+CO_2\)
9,5 → 9,5
\(\rightarrow V_{CO_2}=9,5.22,4=212,8\)
→ hiệu suất phản ứng là
\(\dfrac{159,6}{212,8}.100=75\%\)
\(a.\)
\(m_{CaCO_3}=150\cdot80\%=120\left(g\right)\)
\(n_{CaCO_3}=\dfrac{120}{100}=1.2\left(mol\right)\)
\(CaCO_3\underrightarrow{^{^{t^0}}}CaO+CO_2\)
\(1.2...........1.2\)
\(m_{CaO=}=1.2\cdot56=67.2\left(g\right)\)
\(b.\)
\(n_{CO_2}=\dfrac{27.6}{24}=1.15\left(mol\right)\)
\(n_{CaCO_3}=1.15\left(mol\right)\)
\(m_{CaCO_3}=1.15\cdot100=115\left(g\right)\)
\(m_{TC}=115\cdot20\%=23\left(g\right)\)
a, - Khối lượng CaCO3 trong 150g đá là : 120g
=> \(n_{CaCO3}=\dfrac{m}{M}=1,2\left(mol\right)\)
\(PTHH:CaCO_3\rightarrow CaO+CO_2\)
Theo PTHH : \(n_{CaO}=1,2\left(mol\right)\)
\(\Rightarrow m_{vs}=m_{CaO}=n.M=67,2\left(g\right)\)
b, \(n_{CO2}=\dfrac{V}{24}=1,15\left(mol\right)\)
Theo PTHH : \(n_{CaCO3}=1,15\left(mol\right)\)
\(\Rightarrow m_{CaCO3}=n.M=115\left(g\right)\)
=> %Tạp chất là : \(\left(1-\dfrac{115}{150}\right).100\%=\dfrac{70}{3}\%\)
Vậy ...
a) $CaCO_3 \xrightarrow{t^o} CaO + CO_2$
b) $m_{CaCO_3} = 120 - 120.20\% = 96(gam)$
Theo PTHH :
$n_{CaO} = n_{CaCO_3} = \dfrac{96}{100} = 0,96(mol)$
$\Rightarrow m_{CaO} = 0,96.56 = 53,76(gam)$
c) $n_{CO_2} = n_{CaCO_3} = 0,96(mol)$
$\Rightarrow V_{CO_2} = 0,96.22,4 = 21,504(lít)$
`n_(CaO) = m/M=42/(40+16)=0,75(mol)`
`PTHH: CaCO_3 -> CaO + CO_2`
tỉ lệ 1 ; 1 : 1
n(mol) 0,75<----------0,75--->0,75
`m_(CaCO_3)=nxxM=0,75xx(40+12+16xx3)=75(g)`
Ta có
a, CaCO3 →→ CaO + CO2
=> mCaO + mCO2 = mCaCO3
=> mCaCO3 = 25 ( kg )
b, mđá vôi = 25 : 80 . 100 = 31,25 ( kg )
Bài 3 :
$n_{CaO} = \dfrac{168}{56} = 3(kmol)$
$CaCO_3 \xrightarrow{t^o} CaO + CO_2$
$n_{CaCO_3\ pư} = n_{CaO} = 3(kmol)$
$n_{CaCO_3\ đã\ dùng} = \dfrac{3}{80\%} = 3,75(kmol)$
$m_{CaCO_3} = 3,75.100 = 375(kg)$
$m = \dfrac{375}{80\%} = 468,75(kg)$
Bài 2 :
\(m_{CaCO_3}=280\cdot75\%=210\left(kg\right)\)
\(n_{CaCO_3\left(pư\right)}=\dfrac{210}{100}\cdot80\%=1.68\left(kmol\right)\)
\(CaCO_3\underrightarrow{^{^{t^0}}}CaO+CO_2\)
\(1.68.........1.68......1.68\)
\(m_{CaO}=1.68\cdot56=94.08\left(kg\right)\)
\(V_{CO_2}=1.68\cdot22.4=37.632\left(l\right)=0.037632\left(m^3\right)\)