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a, \(n_{H_2}=\dfrac{22,4}{22,4}=1\left(mol\right)\)
PT: \(2Na+2H_2O\rightarrow2NaOH+H_2\)
Theo PT: \(n_{Na}=n_{NaOH}=2n_{H_2}=2\left(mol\right)\)
\(\Rightarrow m_{Na}=2.23=46\left(g\right)\)
b, \(m_{NaOH}=2.40=80\left(g\right)\)
c, \(n_{HCl}=\dfrac{365}{36,5}=10\left(mol\right)\)
PT: \(NaOH+HCl\rightarrow NaCl+H_2O\)
Xét tỉ lệ: \(\dfrac{2}{1}< \dfrac{10}{1}\), ta được HCl dư.
Theo PT: \(n_{NaCl}=n_{NaOH}=2\left(mol\right)\Rightarrow m_{NaCl}=2.58,5=117\left(g\right)\)
d, \(n_{H_2}=\dfrac{1}{3}\left(mol\right)\)
- Với Fe3O4:
PT: \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)
Theo PT: \(n_{Fe}=\dfrac{3}{4}n_{H_2}=0,25\left(mol\right)\Rightarrow m_{Fe}=0,25.56=14\left(g\right)\)
- Với CuO:
PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Theo PT: \(n_{Cu}=n_{H_2}=\dfrac{1}{3}\left(mol\right)\Rightarrow m_{Cu}=\dfrac{1}{3}.64=\dfrac{64}{3}\left(g\right)\)
\(n_{CaCO_3}=\dfrac{7}{100}=0,07\left(mol\right)\)
\(n_{HCl}=\dfrac{5,475}{36,5}=0,15\left(mol\right)\)
PTHH: CaCO3 + 2HCl --> CaCl2 + CO2 + H2O
Xét \(\dfrac{n_{CaCO_3}}{1}=0,07< \dfrac{n_{HCl}}{2}=0,075\)
=> HCl dư
Do đó, ta có:
PTHH: CaCO3 + 2HCl --> CaCl2 + CO2 + H2O
_______0,075<---0,15__________________________(mol)
=> \(m_{CaCO_3\left(cầnthêm\right)}=\left(0,075-0,07\right).100=0,5\left(g\right)\)
Giả sử M có hóa trị n.
PT: \(2M+nCl_2\underrightarrow{t^o}2MCl_n\)
\(MCl_n+nNaOH\rightarrow M\left(OH\right)_{n\downarrow}+nNaCl\)
Ta có: \(n_{Cl_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Theo PT: \(n_{MCl_n}=\dfrac{2}{n}n_{Cl_2}=\dfrac{0,6}{n}\left(mol\right)\)
\(n_{M\left(OH\right)_n}=\dfrac{21,4}{M_M+17n}\left(mol\right)\)
Theo PT: \(n_{MCl_n}=n_{M\left(OH\right)_n}\Rightarrow\dfrac{0,6}{n}=\dfrac{21,4}{M_M+17n}\)
\(\Rightarrow M_M=\dfrac{56}{3}n\left(g/mol\right)\)
Với n = 3 thì MM = 56 (g/mol) là tm.
Vậy: M là Fe.
\(a.n_P=\dfrac{6,2}{31}=0,2mol\\ 4P+5O_2\xrightarrow[]{t^0}2P_2O_5\)
\(n_{P_2O_5}=0,2.2:4=0,1mol\\ m_{P_2O_5}=0,1.142=14,2g\)
\(b.n_{P_2O_5}=\dfrac{35,5}{142}=0,25mol\\ n_P=0,25.2=0,5mol\\ m_P=0,5.31=15,5g\\ n_{O_2}=\dfrac{0,25.5}{2}=0,625mol\\ V_{O_2}=0,625.24,79=15,49375l\)
a) 2Na + 2HCl ===> 2NaCl + H2
b) nNa = \(\frac{13,8}{23}=0,6\left(mol\right)\)
Theo phương trình, ta có: nNa = mNaCl = 0,6 (mol)
=> mNaCl = 0,6 x 58,5 = 35,1 gam
Theo phương trình, ta có: nH2 = 0,3 (mol)
=> mH2 = 0,3 x 2 = 0,6 (gam)
BTKL: \(m_{K_2O}=m_K+m_{O_2}=11,7+6,4=18,1\left(g\right)\)
\(a.Zn+2HCl\rightarrow ZnCl_2+H_2\)
b. Theo ĐLBTKL, ta có:
\(m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\\ b.\Leftrightarrow6,5+7,3=13,6+m_{H_2}\\ \Leftrightarrow m_{H_2}=\left(6,5+7,3\right)-13,6=0,2\left(g\right)\)
Chúc em học tốt!
nNa=2,3 : 23=0,1 ( mol )
PTHH : 2Na+Cl2-->2NaCl
2 2
0,1 --> 0,1
=> nNaCl=0,1 mol
=>mNaCl=0,1x58,5=5,85 (g)