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PTHH : CaCO3 → CaO + CO2
Áp dụng định luật bảo toàn khối lượng : mCaCO3 = mCaO + mCO2
=> 100 = 56 + mCO2
=> mCO2 = 100 - 56 = 44 gam
1) \(m_{CO_2}=m_{rắn\left(trcpư\right)}-m_{rắn\left(saupư\right)}=100-64,8=35,2\left(g\right)\)
=> \(n_{CO_2}=\dfrac{35,2}{44}=0,8\left(mol\right)\)
=> \(V_{CO_2}=0,8.22,4=17,92\left(l\right)\)
2)
PTHH: CaCO3 --to--> CaO + CO2
0,8<---------0,8<---0,8
=> \(m_{CaCO_3\left(pư\right)}=0,8.100=80\left(g\right)\)
3)
\(m_{CaCO_3\left(bd\right)}=\dfrac{100.90}{100}=90\left(g\right)\)
=> Rắn sau pư chứa CaCO3, CaO, tạp chất
\(m_{tạp.chất}=100-90=10\left(g\right)\)
\(m_{CaCO_3\left(saupư\right)}=90-80=10\left(g\right)\)
\(m_{CaO}=0,8.56=44,8\left(g\right)\)
\(1,n_{CaCO_3}=\dfrac{90\%.100}{100}=0,9\left(mol\right)\\ PTHH:CaCO_3\rightarrow\left(t^o\right)CaO+CO_2\\ Đặt:n_{CaCO_3\left(p.ứ\right)}=a\left(mol\right)\left(a>0\right)\\ Ta.có:m_{rắn}=64,8\left(g\right)\\ \Leftrightarrow10+\left(90-100a\right)+56a=64,8\\ \Leftrightarrow a=0,8\left(mol\right)\\ n_{CO_2}=n_{CaO}=n_{CaCO_3\left(p.ứ\right)}=0,8\left(mol\right)\\ V_{CO_2\left(đktc\right)}=0,8.22,4=17,92\left(l\right)\\ 2,m_{CaCO_3\left(p.ứ\right)}=0,8.100=80\left(g\right)\\ 3,Rắn.sau.nung:m_{tạp.chất}=10\%.100=10\left(g\right)\\ m_{CaO}=0,8.56=44,8\left(g\right)\\ m_{CaCO_3\left(dư\right)}=\left(0,9-0,8\right).100=10\left(g\right)\)
Theo ĐLBT KL, có: mCaCO3 = mCaO + mCO2
⇒ mCO2 = 10 - 5,6 = 4,4 (tấn)
\(a.\)
\(m_{CaCO_3}=150\cdot80\%=120\left(g\right)\)
\(n_{CaCO_3}=\dfrac{120}{100}=1.2\left(mol\right)\)
\(CaCO_3\underrightarrow{^{^{t^0}}}CaO+CO_2\)
\(1.2...........1.2\)
\(m_{CaO=}=1.2\cdot56=67.2\left(g\right)\)
\(b.\)
\(n_{CO_2}=\dfrac{27.6}{24}=1.15\left(mol\right)\)
\(n_{CaCO_3}=1.15\left(mol\right)\)
\(m_{CaCO_3}=1.15\cdot100=115\left(g\right)\)
\(m_{TC}=115\cdot20\%=23\left(g\right)\)
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 ...
\(m_{CaCO_3}=400\cdot90\%=360\left(g\right)\)
\(m_{trơ}=400-360=40\left(g\right)\)
\(n_{CaCO_3}=\dfrac{360}{100}=3.6\left(mol\right)\)
\(a.\)
\(n_{CaCO_3\left(pư\right)}=3.6\cdot75\%=2.7\left(mol\right)\)
\(CaCO_3\underrightarrow{^{^{t^0}}}CaO+CO_2\)
\(2.7........2.7...........2.7\)
\(m_X=m_{CaO}+m_{CaCO_3\left(dư\right)}+m_{trơ}=2.7\cdot56+\left(3.6-2.7\right)\cdot100+40=281.2\left(g\right)\)
\(b.\)
\(\%CaO=\dfrac{2.7\cdot56}{281.2}\cdot100\%=53.77\%\)
\(V_{CO_2}=2.7\cdot22.4=60.48\left(l\right)\)
to
a) Công thức về khối lượng phản ứng:
mCaCO3 = mCaO + mCO2
b) mCaCO3 = 280 + 110 = 390 kg
=> %CaCO3
= \(\frac{390}{560}\) = 69,7%
\(m_{CaCO_3}=90\%.400=360\left(g\right)\\ \rightarrow n_{CaCO_3}=\dfrac{360}{100}=3,6\left(mol\right)\)
PTHH: CaCO3 --to--> CaO + CO2
3,6 ----------> 3,6 -----> 3,6
\(\rightarrow n_{CaO}=3,6.75\%=2,7\left(mol\right)\\ \rightarrow n_{CaCO_3\left(chưa.pư\right)}=3,6-2,7=0,9\left(mol\right)\)
\(\rightarrow m_X=0,9.100+2,7.56=241,2\left(g\right)\\ \%m_{CaO}=\dfrac{0,9.100}{241,2}=37,31\%\)
\(V_Y=V_{CO_2}=3,6.75\%.22,4=60,48\left(l\right)\)
\(m_{CaCO_3}=\dfrac{400\cdot90\%}{100\%}=360g\Rightarrow n_{CaCO_3}=\dfrac{360}{100}=3,6mol\)
\(CaCO_3\underrightarrow{t^o}CaO+CO_2\)
3,6 3,6 3,6
Thực tế: \(n_{CaO}=3,6\cdot75\%=2,7mol\)
\(\Rightarrow m_{CaO}=2,7\cdot56=151,2g\)
Tk:
nCaCO3=100100=1(mol)nCaCO3=100100=1(mol)
nCaO=5656=1(mol)nCaO=5656=1(mol)
PT: CaCO3 to→CaO+CO2to→CaO+CO2
mol 1 1 1
a) mCO2=1.44=44(g)mCO2=1.44=44(g)
VCO2(đktc)=1.22,4=22,4(l)