ta có n_CO2=\(\frac{13.44}{22.4}\)=0,6 mol

bt1) Fe₂O₃+ CO\(\rightarrow\) CO₂+Fe

ta có n_Fe= 0,6 mol

vậy m_Fe=0,6.56=33,6

bt2) ta có n_Fe=1,12:56=0,02 mol

PTHH: Fe203+3CO\(\rightarrow\)2Fe+CO2

                                  0,02\(\rightarrow\)0,01(mol)

V_CO= 0,01. 22,4=0,224(lít)

bạn ơi mink nhầm bài 2 mink làm phải là bài 3 mới đúng bạn nhé

\(n_{O_2}= \dfrac{20-17,12}{32} = 0,09(mol) \)

2KMnO₄ \(\xrightarrow{t^o}\) K₂MnO₄ + MnO₂ + O₂

0,18................0,09.........0,09.....0,09...............(mol)

Sau phản ứng,

\(m_{KMnO_4} = 20-0,18.158 = -8,44 <0\)(Sai đề)

Ta có PTHH: 2KMnO₄ -----> K₂MnO₄ + MnO₂ + O₂.

Ta có: n_KMnO4= 20/158=0,1(mol)

Theo PT, ta có: n_K2MnO4= 1/2 . 0,1=0,05(mol)

=> m_K2MnO4= 0,05.197=9,85(g)

n_MnO2=n_K2MnO4=0,05(mol)

=> m_MnO2=0,05.87=4,35(g)

=> Khối lượng của chất rắn sau phản ứng là: 9,85+4,35=14,2(g)

                                   (Chúc bạn làm tốt nhé )

1) Trong 100 g đá vôi có 90 g CaCO3 (0,9 mol).

Gọi x là số mol CO2, theo pt ta có: 64,8 = 56x + 90 - 100x hay x = 0,572 mol.

V = 0,572.22,4 = 12,829 lít.

2) m = 100x = 57,2 g.

3) mCaCO3 = 90 - 100x = 32,8 g; mCaO = 64,8 - 32,8 = 32g.

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 --t^o--> 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

\(m_{tạp.chất}=100-90=10\left(g\right)\)

\(m_{CaCO_3\left(saupư\right)}=90-80=10\left(g\right)\)

\(m_{CaO}=0,8.56=44,8\left(g\right)\)

\(1,n_{CaCO_3}=\dfrac{90\%.100}{100}=0,9\left(mol\right)\\ PTHH:CaCO_3\rightarrow\left(t^o\right)CaO+CO_2\\ Đặt:n_{CaCO_3\left(p.ứ\right)}=a\left(mol\right)\left(a>0\right)\\ Ta.có:m_{rắn}=64,8\left(g\right)\\ \Leftrightarrow10+\left(90-100a\right)+56a=64,8\\ \Leftrightarrow a=0,8\left(mol\right)\\ n_{CO_2}=n_{CaO}=n_{CaCO_3\left(p.ứ\right)}=0,8\left(mol\right)\\ V_{CO_2\left(đktc\right)}=0,8.22,4=17,92\left(l\right)\\ 2,m_{CaCO_3\left(p.ứ\right)}=0,8.100=80\left(g\right)\\ 3,Rắn.sau.nung:m_{tạp.chất}=10\%.100=10\left(g\right)\\ m_{CaO}=0,8.56=44,8\left(g\right)\\ m_{CaCO_3\left(dư\right)}=\left(0,9-0,8\right).100=10\left(g\right)\)

\(m_{CaCO_3}=90\%.400=360\left(g\right)\\ \rightarrow n_{CaCO_3}=\dfrac{360}{100}=3,6\left(mol\right)\)

PTHH: CaCO₃ --t^o--> 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: \(n_{CaO}=3,6\cdot75\%=2,7mol\)

\(\Rightarrow m_{CaO}=2,7\cdot56=151,2g\)

a) \(n_{CaCO_3\left(bđ\right)}=\dfrac{400.90\%}{100}=3,6\left(mol\right)\)

=> \(n_{CaCO_3\left(pư\right)}=\dfrac{3,6.75}{100}=2,7\left(mol\right)\)

PTHH: CaCO₃ --t^o--> CaO + CO₂

              2,7---------->2,7---->2,7

=> m_X = m_{đá vôi} - m_CO2 = 400 - 2,7.44 = 281,2 (g)

b) 

m_CaO = 2,7.56 = 151,2 (g)

=> \(\%CaO=\dfrac{151,2}{281,2}.100\%=53,77\%\)

V_CO2 = 2,7.22,4 = 60,48 (l)

\(m_{CaCO_3}=400\cdot90\%=360\left(g\right)\)

\(m_{trơ}=400-360=40\left(g\right)\)

\(n_{CaCO_3}=\dfrac{360}{100}=3.6\left(mol\right)\)

\(a.\)

\(n_{CaCO_3\left(pư\right)}=3.6\cdot75\%=2.7\left(mol\right)\)

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

\(2.7........2.7...........2.7\)

\(m_X=m_{CaO}+m_{CaCO_3\left(dư\right)}+m_{trơ}=2.7\cdot56+\left(3.6-2.7\right)\cdot100+40=281.2\left(g\right)\)

\(b.\)

\(\%CaO=\dfrac{2.7\cdot56}{281.2}\cdot100\%=53.77\%\)

\(V_{CO_2}=2.7\cdot22.4=60.48\left(l\right)\)

\(Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\uparrow\left(1\right)\\ CaO+H_2O\rightarrow Ca\left(OH\right)_2\left(2\right)\\ n_{H_2}=\dfrac{V_{H_2\left(đktc\right)}}{22,4}=\dfrac{3,36}{22,4}=0,15\left(l\right)\\ \Rightarrow n_{Ca}=n_{Ca\left(OH\right)_2\left(1\right)}=n_{H_2}=0,15\left(mol\right)\\ \Rightarrow n_{CaO}=\dfrac{11,6-0,15.40}{56}=0,1\left(mol\right)\\ \Rightarrow n_{Ca\left(OH\right)_2\left(2\right)}=n_{CaO}=0,1\left(mol\right)\\ \Rightarrow n_{Ca\left(OH\right)_2\left(tổng\right)}=n_{Ca\left(OH\right)_2\left(1\right)}+n_{Ca\left(OH\right)_2\left(2\right)}=0,15+0,1=0,25\left(mol\right)\\ \Rightarrow m_{rắn}=m_{Ca\left(OH\right)_2\left(tổng\right)}=0,25.74=18,5\left(g\right)\)