Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
C + O2---t*--> CO2
CO2+ CaO--->CaCO3
CaCO3----t*--->CaO+CO2
2NaOH+CO2----->Na2CO3+H2O
Na2CO3 + Ba(OH)2---->2NaOH + BaCO3
2NaOH + Al2O3----->2NaAlO2 + H2O
2Ca + O2 ----.>2CaO
CaO + H2O----->Ca(OH)2
Ca(OH)2 + CO2----->CaCO3 + H2O
CaCO3 + BaCl2----->CaCl2 + BaCO3
Chi?
4Fe+3O2---t*--->2Fe2O3
Fe2O3+ 6HCl------>2FeCl3+3H2O
FeCl3+Al(OH)3------>AlCl3 + Fe(OH)3
AlCl3+3NaOH------>Al(OH)3+3NaCl
2Al(OH)3 ----t*---->Al2O3 + 3H2O
bạn tự cân bằng nhé chứ ko lâu lắm
a) \(S+O_2-to>SO_2\)
\(SO_2+O_2-\frac{to}{xt}>SO_3\)
\(SO_3+H_2O->H_2SO_4\)
\(NaOH+H_2SO_4->Na_2SO_4+H_2O\)
\(BaCl_2+Na_2SO_4->BaSO_4+NaCl\)
\(\left(1\right)H_2S+2NaOH\rightarrow2H_2O+Na_2S\)
\(\left(2\right)Na_2S+CuSO_4\rightarrow Na_2SO_4+CuS\)
\(\left(3\right)8Na+10HNO_3\rightarrow3NaNO_3+NH_4NO_3+8H_2O\)
\(\left(4\right)Na_2O+N_2O_5\rightarrow2NaNO_3\)
\(\left(5\right)Na_2O\)
\(\left(6\right)NaHCO_3+Ca\left(OH\right)_2\rightarrow NaOH+H_2O+CaCO_3\)
\(\left(7\right)NaOH+CO_2\rightarrow NaHCO_3\)
\(\left(8\right)2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)
\(\left(9\right)Na_2CO_3+Ca\left(OH\right)_2\rightarrow2NaOH+CaCO_3\)
\(\left(10\right)2NaOH+SO_3\rightarrow Na_2SO_4+H_2O\)
\(\left(11\right)Na_2SO_4+Ba\left(OH\right)_2\rightarrow BaSO_4+2NaOH\)
\(\left(12\right)NaOH+N_2O_5\rightarrow2NaNO_3\)
\(\left(13\right)Mg\left(HCO_3\right)_2\)
\(\left(14\right)MgSO_4\)
\(\left(15\right)Mg\left(HCO_3\right)_2+2HCl\rightarrow MgCl_2+CO_2+H_2O\)
\(\left(16\right)MgO+2HCl\rightarrow MgCl_2+H_2O\)
1, CaCO3 + C-->CaO + 2CO
2,CaO+H2O-->Ca(OH)2
3,Ca(OH)2+2HNO3-->Ca(NO3)2 +2H2O
4, 3Ca(NO3)2 + 2(NH4)3PO4 --> Ca3(PO4)2 + 6NH4NO3
5, Hình như bạn viết sai CTHH rồi hay sao ý:<<
Câu 2:
nCO2= 0,5 mol
PTHH: CO2+Ca(OH)2 --> CaCO3 +H2o
Theo phương trình: nCa(OH)2 dư= nCO2= 0.5 mol
nCaCO3= nCO2= 0.5 mol
-> mCa(OH)2= 0,5 . 74=37g
mCaCO3 dư= 0,5 . 100= 50g
Bài 1:
1) CaCO3 → CaO + CO2
2) CaO + H2O → Ca(OH)2
3) Ca(OH)2 + N2O5 → Ca(NO3)2 + H2O
4) 3Ca(NO3)2 + 2Na3PO4 → Ca3(PO4)2 + 6NaNO3
5) sai đề thì phải
Bài 2:
CO2 + Ca(OH)2 → CaCO3↓ + H2O
\(n_{CO_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
Theo PT: \(n_{CaCO_3}=n_{CO_2}=0,5\left(mol\right)\)
\(\Rightarrow m_{CaCO_3}=0,5\times100=50\left(g\right)\)
CaCo3-> CaO+CO2
CaO + H2O->Ca(OH)2
Ca(OH)2 + CO2 -> CaCO3 + H2O
CaCO3 + 2HCl ->CaCl2 + H2O+ CO2
CaCl2 + Na2CO3 -> CaCO3 + 2NaCl
CaCO3 \(\underrightarrow{t^o}\) CaO + CO2
CaO + H2O \(\rightarrow\) Ca(OH)2
Ca(OH)2 + CO2 \(\rightarrow\) CaCO3 + H2O
CaCO3 + 2HCl \(\rightarrow\) CaCl2 + H2O + CO2
CaCl2 + Na2CO3 \(\rightarrow\) CaCO3 + 2NaCl
4Na+O2----->2Na2O
Na2O+H2O--->2NaOH
2NaOH dư+CO2----->Na2CO3+H2O
Na2CO3+CaCl2------>CaCO3+2NaCl
CaCO3---t0---->CaO+CO2
CaO+H2O------>Ca(OH)2
Ca(OH)2+CO2 dư----->Ca(HCO3)2
-)2Na+2H2O----->2NaOH+H2
FeO+2NaOH----->Na2O+Fe(OH)2
Na2O+CO2------->Na2CO3
nNaOH=1,2(mol)
Ta có tỉ lệ :
nNaOH/nCO2=1,6⇒ Tạo hai muối.
CO2+2NaOH⇀Na2CO3+H2O
CO2+NaOH⇀NaHCO3
ta có hệ pt
⇒2x+y=1,2
x+y=0,75
⇒x=0,45;y=0,3
mNa2CO3=n.M=0,3.84=25,2(g)
mNaHCO3=n.M=0,45.106=47,7(g)
mA=25,2+47,7=72,9(g)
b,Na2CO3+BaCl2−−>BaCO3+2NaCl
0.45--------------------->0.45mol
mBaCO3BaCO3=197.0.45=88.65g
a) CaCO3------> CaO+ CO2
CaO+ H2O------>Ca(OH)2
Ca(OH)2+ CO2------> CaCO3+ H2O
b) Ca(OH)2+ FeSO4-----> Fe(OH)2↓+ CaSO4↓
2Fe(OH)2+ 1/2O2+ H2O-----> 2Fe(OH)3
2Fe(OH)3----to--> Fe2O3+ 3H2O
Fe2O3+ 2Al-----to----> 2Fe+ Al2O3
c) Ca(OH)2+ 2CO2----> Ca(HCO3)2
Ca(HCO3)2----to-->CaCO3+ CO2+ H2O
1)
NaHSO4 + Na2CO3 --> Na2SO4 + CO2 + H2O
2NaHSO4 + Na2CO3 --> 2Na2SO4 + CO2 + H2O
2NaHSO4 + BaCl2 --> BaSO4 + Na2SO4 + 2HCl
2NaHSO4 + Ba(HCO3)2 --> Na2SO4 + BaSO4 + 2CO2 + 2H2O
2NaHSO4 + Na2S --> 2Na2SO4 + H2S
2)
a) Na2SO4 + Ba(NO3)2 ( X1) --> BaSO4 + 2NaNO3 (Y1)
b) Ca(HCO3)2 + Na2CO3(X2) --> CaCO3 + 2NaHCO3 (Y2)
c) CuSO4 + H2S (X3) --> CuS + H2SO4 (Y3)
d) 3MgCl2 + 3Na2PO4 (X4) --> Mg3(PO4)2 + 6NaCl (Y4)
a) Na2SO4 + BaCl2 → BaSO4↓ + 2NaCl
X1: BaCl2
Y1:NaCl
b) Ca(HCO3)2 + Ca(OH)2 → 2CaCO3↓ + 2H2O
X2: Ca(OH)2
Y2: H2O
c) CuSO4 + Na2S → CuS↓ + Na2SO4
X3: Na2S
Y3: Na2SO4
d) 3MgCl2 + 2Na3PO4 → Mg3(PO4)2↓ + 6NaCl
X4: Na3PO4
Y4: NaCl
\(\text{2Na+2H2O}\rightarrow\text{2NaOH+H2↑}\)
\(\text{2NaOH(đậm đặc)+CO2}\rightarrow\text{Na2CO3+H2O}\)
\(\left\{{}\begin{matrix}\text{CO2+NaOH}\rightarrow\text{NaHCO3}\\\text{H2CO3+NaOH}\rightarrow\text{NaHCO3+H2O}\end{matrix}\right.\)
\(\text{Na2CO3+Ca(OH)2}\rightarrow\text{CaCO3↓+2NaOH}\)
\(\text{CaCO3 + 2HCl → H2O + CO2 + CaCl2}\)
\(\text{CaCl2+Na2CO3}\rightarrow\text{CaCO3↓+2NaCl}\)
\(\text{CaCO3(thể rắn)+CO2+H2O}\rightarrow\text{Ca(HCO3)2(dung dịch pha loãng)}\)
\(\text{Ca(HCO3)2 → CaCO3 + H2O + CO2}\)
\(\text{Ca(HCO3)2 + 2HCl → 2H2O + 2CO2 + CaCl2}\)
\(\text{2Li3PO4+3CaCl2(đậm đặc)}\rightarrow\text{6LiCl↓+Ca3(PO4)2↓}\)
\(\text{NaHCO3+HCl}\rightarrow\text{NaCl+CO2↑+H2O}\)
\(\text{2NaCl+2H2O}\rightarrow\text{H2↑+2NaOH+Cl2↑}\)