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\(n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{2,4}{22,4}\approx0,11\left(mol\right)\)
\(n_{O_2}=\dfrac{V_{O_2}}{22,4}=\dfrac{1,6}{22,4}\approx0,07\)
\(2H_2+O_2\rightarrow2H_2O\)
2 mol-1mol---2 mol
Ta có: \(\dfrac{n_{H_2}}{2}=\dfrac{0,11}{2}\)
\(\dfrac{n_{O_2}}{1}=\dfrac{0,07}{1}\)
\(\Rightarrow\dfrac{n_{H_2}}{2}< \dfrac{n_{O_2}}{1}\)
Vậy \(O_2\) dư
Số mol O2 dư:
\(n_{O_2}=\dfrac{0,07.1}{2}=0,035\left(mol\right)\)
Khối lượng O2 dư
\(m_{O_2}=0,035.32=1,12\left(g\right)\)
Khối lượng nước thu được:
\(n_{H_2O}=\dfrac{0,07.2}{2}=0,07\left(mol\right)\)
\(\Rightarrow m_{H_2O}=n_{H_2O}.M_{H_2O}=0,07.18=1,26\left(g\right)\)
\(n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{4,48}{22,4}=0,2mol\)
\(n_{Fe_2O_3}=\dfrac{m_{Fe_2O_3}}{M_{Fe_2O_3}}=\dfrac{23,2}{160}=0,145mol\)
\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)
0,145 > 0,2 ( mol )
1/15 0,2 2/15 ( mol 0
Chất còn dư là \(Fe_2O_3\)
\(m_{Fe_2O_3\left(du\right)}=n_{Fe_2O_3\left(du\right)}.M_{Fe_2O_3}=\left(0,145-\dfrac{1}{15}\right).160=12,53g\)
\(m_{Fe}=n_{Fe}.M_{Fe}=\dfrac{2}{15}.56=7,4666g\)
nNa = 4.6/23 = 0.2 (mol)
Na + H2O => NaOH + 1/2H2
0.2....................0.2..........0.1
VH2 = 0.1*22.4 = 2.24 (l)
mNaOH = 0.2*40 = 8 (g)
Đề thiếu khối lượng nước rồi em nhé !
\(a) Zn + H_2SO_4 \to ZnSO_4 + H_2\\ n_{Zn} = \dfrac{13}{65} = 0,2 < n_{H_2SO_4} = \dfrac{200.24,5\%}{98} = 0,5 \to H_2SO_4\ dư\\ n_{H_2SO_4\ pư} =n_{Zn} = 0,2(mol)\\ \Rightarrow m_{H_2SO_4\ dư} = (0,5 - 0,2).98 = 29,4(gam)\\ c) n_{FeSO_4} = n_{H_2} = n_{Zn} = 0,2(mol)\\ m_{FeSO_4} = 0,2.152 = 30,4(gam)\\ V_{H_2} = 0,2.22,4 = 4,48(lít)\)
a, \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
\(m_{HCl}=36,5.15\%=5,475\left(g\right)\Rightarrow n_{HCl}=\dfrac{5,475}{36,5}=0,15\left(mol\right)\)
PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,15}{2}\), ta được Mg dư.
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,075\left(mol\right)\Rightarrow V_{H_2}=0,075.22,4=1,68\left(l\right)\)
b, \(n_{Mg\left(pư\right)}=\dfrac{1}{2}n_{HCl}=0,075\left(mol\right)\Rightarrow n_{Mg\left(dư\right)}=0,1-0,075=0,025\left(mol\right)\)
\(\Rightarrow m_{Mg\left(dư\right)}=0,025.24=0,6\left(g\right)\)
c, - Cách 1:
\(n_{MgCl_2}=\dfrac{1}{2}n_{HCl}=0,075\left(mol\right)\Rightarrow m_{MgCl_2}=0,075.95=7,125\left(g\right)\)
- Cách 2:
Theo ĐLBT KL, có: mMg (pư) + mHCl = mMgCl2 + mH2
⇒ mMgCl2 = 2,4 - 0,6 + 5,475 - 0,075.2 = 7,125 (g)
\(n_K=\dfrac{3,8}{39}=\dfrac{19}{195}mol\)
\(n_{H_2O}=\dfrac{101,8}{18}=\dfrac{509}{90}mol\)
\(2K+2H_2O\rightarrow2KOH+H_2\)
19/195 < 509/90 ( mol )
19/195 19/195 19/195 ( mol )
Chất dư là H2O
\(m_{H_2O\left(dư\right)}=\left(\dfrac{509}{90}-\dfrac{19}{195}\right).18\approx100,04g\)
\(m_{KOH}=\dfrac{19}{195}.56\approx5,45g\)
\(a,n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\\ n_{H_2O}=\dfrac{5,4}{18}=0,3\left(mol\right)\)
PTHH: 2Na + 2H2O ---> 2NaOH + H2
LTL: 0,2 < 0,3 => H2O dư
Theo pthh: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{1}{2}n_{Na}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\\n_{NaOH}=n_{H_2}=0,2\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}b,V_{H_2}=0,1.22,4=2,24\left(l\right)\\m_{NaOH}=0,2.40=8\left(g\right)\end{matrix}\right.\)
Ta có:
\(n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)
\(n_{H_2O}=\dfrac{3,6}{18}=0,2\left(mol\right)\left(phần.này.mik.sửa.lại.đề\right)\)
a. \(PTHH:Na_2O+H_2O--->2NaOH\)
Ta thấy: \(\dfrac{0,1}{1}< \dfrac{0,2}{1}\)
Vậy H2O dư.
Theo PT: \(n_{NaOH}=2.n_{Na_2O}=2.0,1=0,2\left(mol\right)\)
\(\Rightarrow m_{NaOH}=40.0,2=8\left(g\right)\)
b. Ta có: \(n_{H_2O_{PỨ}}=n_{Na_2O}=0,1\left(mol\right)\)
\(\Rightarrow n_{H_2O_{dư}}=0,2-0,1=0,1\left(mol\right)\)
\(\Rightarrow m_{H_2O_{dư}}=0,1.18=1,8\left(g\right)\)
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