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\(n_{Na}=0.02\left(mol\right)\)
\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)
\(0.02....................0.02........0.01\)
\(V_{H_2}=0.01\cdot22.4=0.224\left(l\right)\)
\(m_{NaOH}=0.02\cdot40=0.8\left(g\right)\)
\(C\%_{NaOH}=\dfrac{0.8}{0.46+200-0.01\cdot2}\cdot100\%=0.4\%\)
\(n_{Ca} = a(mol)\\ Ca + 2H_2O \to Ca(OH)_2 + H_2\\ n_{H_2} = n_{Ca(OH)_2} = n_{Ca} = a(mol)\\ m_{dd\ sau\ pư} = 40a + 200 - 2a = 200 + 38a(gam)\\ C\%_{Ca(OH)_2} = \dfrac{74a + 200.1\%}{200 + 38a}.100\% = 2\%\\ \Rightarrow a = \dfrac{50}{1831} \to m_{Ca} = \dfrac{2000}{1831} =1,09(gam)\)
Ta có: \(C\%=20\%=\dfrac{m_{NaCl}}{250}.100\%\)
\(\Rightarrow m_{NaCl}=50\left(g\right)\)
Ta có \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(m_{NaOH}=100.16\%=16\left(g\right)\Rightarrow n_{NaOH}=\dfrac{16}{40}=0,4\left(mol\right)\)
PT: \(2Na+2H_2O\rightarrow2NaOH+H_2\)
\(Na_2O+H_2O\rightarrow2NaOH\)
Theo PT: \(n_{Na}=2n_{H_2}=0,2\left(mol\right)\)
\(n_{NaOH}=n_{Na}+2n_{Na_2O}\Rightarrow n_{Na_2O}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Na}=\dfrac{0,2.23}{0,2.23+0,1.62}.100\%\approx42,6\%\\\%m_{Na_2O}\approx57,4\%\end{matrix}\right.\)
1) Ta có :m CuSO4=200.8%=16 gam\(\rightarrow\)mH2O=200-16=184 gam
\(\rightarrow\)Trong dung dịch CuSO4 2% chứa 98% H2O
\(\rightarrow\)mCuSO4 trong CuSO4 2%=\(\frac{\text{184.2%}}{98\%}\)=3,755 gam
\(\rightarrow\) mCuSO4 trong CuSO4.5H2O=16-3,755=12,245 gam
\(\rightarrow\) Trong 1 mol CuSO4.5H2O chứa 1 mol CuSO4
\(\rightarrow\) 250 gam CuSO4.5H2O chứa 160 gam CuSO4
\(\rightarrow\) mCuSO4.5H2O=\(\frac{\text{12,245.250}}{160}\)=19,133 gam
2)
N2O5 + H2O\(\rightarrow\) 2HNO3
Ta có: mHNO3=200.20%=40 gam
Gọi số mol N2O5 là a; HNO3 trong dung dịch ban đầu là b
\(\rightarrow\) 63(2a+b)=40
Mặt khác: m N2O5=108a; m HNO3 trong dung dịch=63b -> m dung dịch =630b
\(\rightarrow\) 108a+630b=200
Giải được: a=\(\frac{25}{144}\); b=\(\frac{145}{504}\)
\(\rightarrow\) mN2O5=18,75 gam; m dd HNO3=181,25 gam
3)
Gọi khối lượng của dung dịch H2SO4 10% và H2SO4 20% lần lượt là a; b
\(\rightarrow\) BTKL: a+b=200
Ta có: mH2SO4=200.16%=32 gam=10%.a+20%b
Giải được: a=80; b=120
\(n_{Na}=\dfrac{13,8}{23}=0,6\left(mol\right)\\ pthh:Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)
0,6 0,6 0,3
\(V_{H_2}=0,3.22,4=6,72\left(l\right)\\ c,m_{\text{dd}}=13,8+286,8-\left(0,3.2\right)=300\left(g\right)\\ C\%=\dfrac{0,6.40}{300}.100\%=8\%\)
\(n_{Na}\) = \(\dfrac{13,8}{23}\) = 0,6 mol
Theo PTHH:
a) \(2Na+2H_2O\underrightarrow{t^o}2NaOH+H_2\)
2 2 2 1 (mol)
0,6 \(\rightarrow\) 0,6 \(\rightarrow\) 0,6 \(\rightarrow\) 0,3 (mol)
b) \(V_{H_2}\) = 0,3.22,4 = 6,72l
c) \(m_{dd}\) = 13,8 + 286,8 - 0,3.2 = 300g
\(C\%\) = \(\dfrac{0,6.40}{300}\).100% = 8%
\(m_{H_2O}=37,6.1=37,6\left(g\right)\\ m_{ddNaOH}=12,4+37,6=50\left(g\right)\\ C\%_{ddNaOH}=\dfrac{12,4}{50}.100=24,8\%\)
Ta có: \(D_{H_2O}=1\left(g/mol\right)\Rightarrow m_{H_2O}=37,6.1=37,6\left(g\right)\)
\(\Rightarrow C\%_{ddNaOH}=\dfrac{12,4.100\%}{37,6}=32,98\%\)
1.
P2O5 + 3H2O \(\rightarrow\) 2H3PO4
Ta có: m H3PO4=200.29,4%=58,8 gam
\(\rightarrow\)nH3PO4=\(\frac{58,8}{\text{3+31+16.4}}\)=0,6 mol\(\rightarrow\)nP2O5=0,3 mol
\(\rightarrow\) mP2O5=42,6 gam \(\rightarrow\)mH2O=200-42,6=157,4 gam
2) Ta có : mH2SO4=2000.19,6%=392 gam
\(\rightarrow\)nH2SO4= 4 mol
H2SO4.3SO3 + 3H2O\(\rightarrow\) 4H2SO4
\(\rightarrow\) nH2SO4.3SO3\(\frac{1}{4}\)nH2SO4=1 mol
\(\rightarrow\) mH2SO4.3SO3=338 gam
\(\rightarrow\)mH2O=2000-338=1662 gam
3)
Ba + 2H2O\(\rightarrow\) Ba(OH)2 + H2
Ta có: mBa(OH)2=200.17,1%=34,2 gam
\(\rightarrow\)nBa(OH)2=\(\frac{34,2}{\text{137+17.2}}\)=0,2 mol =nBa=nH2
\(\rightarrow\)mBa=0,2.137=27,4 gam
BTKL: mBa + mH2O= m dung dịch + mH2
\(\rightarrow\) 27,4+mH2O=200+0,2.2
\(\rightarrow\) mH2O=173 gam
nNaOH=\(\frac{400.12\%}{100\%.40}\)=1,2(mol)
m\(_{ddNaOH\left(20\%\right)}\)=\(\frac{100\%.40.1,2}{20\%}\)=240(g)
m\(_{H2O}\)cần dùng =400-240=160(g)
cho mk hỏi chút 40 là gì zậy?