1)

1,2 tấn = 1200(kg)

5 tạ = 500(kg)

\(m_{CaCO_3} = 1200.80\% = 960(kg)\)

\(CaCO_3 \xrightarrow{t^o} CaO + CO_2\\ n_{CaCO_3\ pư} = n_{CaO} = \dfrac{500}{56}(mol)\\ \Rightarrow H = \dfrac{\dfrac{500}{56}.100}{960}.100\% = 93\%\)

1 (H)= 93,11%

2 (H)=88.08%

m cao=1.064(tấn)

==> m cr = 1.065(tấn)

%m cao = 56%

mCaCO3 = 300*0.8 = 240 (kg) 

nCaCO3 = 240*1000/100 = 2400 (mol) 

CaCO3 -to-> CaO + CO2 

2400_______2400___2400 

mCaO = 2400*72 = 172800 (g) = 172.8 (kg) 

VCO2 = 2400*22.4 = 53760 (l) 

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

nCaO = 33,6/56 = 0,6 (mol)

nCO₂ = 26,4/44 = 0,6 (mol)

=> nCaCO₃ = 0,6 mol <=> mCaCO₃ = 0,6.100 = 60 (g)

m_{CaCO3(nguyên chất)} = 1 x 90% = 0,9 tấn

PTHH:

CaCO3 ==(nhiệt)==> CaO + CO₂

1mol.............................1mol

100 tấn...........................56 tấn

0,9 tấn........................ ......x tấn

=> mCaO(theo phương trình) = \(\frac{0,9.56}{100}=0,504\) tấn

Mà thực tế thu được 0,45 tấn

=> H = \(\frac{0,45}{0,504}.100\%=89,3\%\)

CaCO3(562507)→CaO(562507)+CO2CaCO3(562507)→CaO(562507)+CO2

mCaCO3=1000000.90%=900000(g)mCaCO3=1000000.90%=900000(g)

nCaO=45000056=562507nCaO=45000056=562507

⇒mCaCO3(pứ)=562507.100=56250007(g)⇒mCaCO3(pứ)=562507.100=56250007(g)

⇒H=56250007900000.100%=89,29%

\(a.\)

\(m_{CaCO_3}=150\cdot80\%=120\left(g\right)\)

\(n_{CaCO_3}=\dfrac{120}{100}=1.2\left(mol\right)\)

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

\(1.2...........1.2\)

\(m_{CaO=}=1.2\cdot56=67.2\left(g\right)\)

\(b.\)

\(n_{CO_2}=\dfrac{27.6}{24}=1.15\left(mol\right)\)

\(n_{CaCO_3}=1.15\left(mol\right)\)

\(m_{CaCO_3}=1.15\cdot100=115\left(g\right)\)

\(m_{TC}=115\cdot20\%=23\left(g\right)\)

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)

\(m_{CaCO_3} = 250.1000.75\% = 187500(kg)\\ \Rightarrow n_{CaCO_3} = \dfrac{187500}{100} = 1875(kmol)\\ CaCO_3 \xrightarrow{t^o} CaO + CO_2\)

Theo PTHH : \(n_{CaO} = n_{CaCO_3} = 1875\ mol\\ \Rightarrow m_{CaO} = 1875.56 = 105000(kg)\)

b)

\(n_{CO_2} = n_{CaCO_3} = 1875\ mol\\ \Rightarrow V_{CO_2} = 1875.22,4 = 42000(lít)\)

Ta có

a, CaCO₃  CaO + CO₂

=> m_CaO + m_CO2 = m_CaCO3

=> m_CaCO3 = 25 ( kg )

b, m_{đá vôi} = 25 : 80 . 100 = 31,25 ( kg )

1)

1,2 tấn = 1200(kg)

5 tạ = 500(kg)

mCaCO3=1200.80%=960(kg)mCaCO3=1200.80%=960(kg)

\(m_{\text{CaCO_3}}=1000.95\%=950kg\\ \rightarrow n_{\text{CaCO_3}}=9,5mol\)

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

9,5            →   9,5

\(\rightarrow V_{CO_2}=9,5.22,4=212,8\)

→  hiệu suất phản ứng là

\(\dfrac{159,6}{212,8}.100=75\%\)

\(m_{CO_2} = m_{giảm} = 50.22\% = 11(gam)\\ CaCO_3 \xrightarrow{t^o} CaO + CO_2\\ n_{CaCO_3\ pư} =n_{CO_2} = \dfrac{11}{44} = 0,25(mol)\\ \Rightarrow m_{CaCO_3\ bị\ phân\ hủy} = 0,25.100 = 25(gam)\)

Ta có: m_CaCO3 = 500.80% = 400 (g)

m _{chất rắn} = 400.78% = 312 (g)

Theo ĐLBT KL, có: m_CO2 = 400 - 312 = 88 (g)

\(\Rightarrow n_{CO_2}=\dfrac{88}{44}=2\left(mol\right)\)

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

_____2_____________2 (mol)

\(\Rightarrow m_{CaCO_3\left(pư\right)}=2.100=200\left(g\right)\)

Bạn tham khảo nhé!