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\(n_{CuSO4}=\dfrac{16\%.50}{100\%.160}=0,05\left(mol\right)\)
Pt : \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
0,05-------->0,1---------->0,05--------->0,05
a) \(C\%_{ddNaOH}=\dfrac{0,1.40}{250}.100\%=1,6\%\)
b) \(m_{ddspu}=50+250-0,05.98=295,1\left(g\right)\)
\(C\%_{Na2SO4}=\dfrac{0,05.142}{295,1}.100\%=2,41\%\)
\(n_{CuSO_4}=\dfrac{m_{dd}\cdot C\%}{100\cdot M}=\dfrac{50\cdot16\%}{100\cdot\left(64+32+16\cdot4\right)}=0,05\left(mol\right)\)
\(PTHH:CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
1 2 1 1
0,05 0,1 0,05 0,05 (mol)
\(a)C\%_{NaOH}=\dfrac{n\cdot100\cdot M}{m_{dd}}=\dfrac{01\cdot100\cdot\left(23+16+1\right)}{250}=1,6\%\)
\(b)m_{dd-sau-pư}=m_{dd_đ}+m_{ct_đ}-m\downarrow-m\uparrow\)
\(=50+250-\left(0,05\cdot23+32+16\cdot4\right)=294,05\left(g\right)\)
\(C\%_{Na_2SO_4}=\dfrac{n\cdot100\cdot M}{m_{dd}}=\dfrac{0,05\cdot100\cdot\left(23\cdot2+32+16\cdot4\right)}{294,05}\approx2,41\%.\)
\(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
a) Pt : \(CuO+2HCl\rightarrow CuCl_2+H_2O|\)
1 2 1 1
0,1 0,2 0,1
b) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
⇒ \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(C_{ddHCl}=\dfrac{7,3.100}{200}=3,65\)0/0
c) \(n_{CuCl2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)
⇒ \(m_{CuCl2}=0,1.135=13,5\left(g\right)\)
Chúc bạn học tốt
a)mH2SO4=\(\dfrac{200.7,3\text{%}}{100\%}\)=14,6g
nHCl=\(\dfrac{14,6}{36,5}\)=0,4(mol)
PTHH:
NaOH+ HCl→ NaCl+ H2O
1 1 1 1
0,4 0,4 0,4 (mol)
⇒mNaOH=0,4.40=16(g)
Nồng độ % của dd NaOH cần dùng là:
C%NaOH=\(\dfrac{16}{200}\) .100%=8%
b)Ta có:mdd spứ=mdd trc pứ=400g
mNaCl=0,4.58,5=23,4g
Nồng độ % dd muối tạo thành sau pứ là:
C%dd NaCl=\(\dfrac{23,4}{400}\) .100%=5,85%
a, \(CuO+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Cu+H_2O\)
b, \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
Theo PT: \(n_{CH_3COOH}=2n_{CuO}=0,2\left(mol\right)\Rightarrow C_{M_{CH_3COOH}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
c, \(n_{\left(CH_3COO\right)_2Cu}=n_{Cu}=0,1\left(mol\right)\Rightarrow m_{\left(CH_3COO\right)_2Cu}=0,1.182=18,2\left(g\right)\)
\(a,PTHH:Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\\ b,n_{Ca\left(OH\right)_2}=\dfrac{7,4}{74}=0,1\left(mol\right)\\ \Rightarrow n_{HCl}=2n_{Ca\left(OH\right)_2}=0,2\left(mol\right)\\ \Rightarrow m_{CT_{HCl}}=0,2\cdot36,5=7,3\left(g\right)\\ \Rightarrow C\%_{HCl}=\dfrac{7,3}{200}\cdot100\%=3,65\%\\ c,CaCl_2+H_2SO_4\rightarrow CaSO_4+2HCl\\ n_{H_2SO_4}=1\cdot0,25=0,25\left(mol\right)\\ n_{CaCl_2}=n_{Ca\left(OH\right)_2}=0,1\left(mol\right)\)
Do đó sau p/ứ H2SO4 dư
\(\Rightarrow n_{CaSO_4}=n_{CaCl_2}=0,1\left(mol\right)\\ \Rightarrow m_{CaSO_4}=0,1\cdot136=13,6\left(g\right)\)
Ca(OH)2+ H2CL-> CaCL2+ H2O
số n của Ca(OH)2 là :
A) nCa(OH)2 =m/M=7,4/74=0,1 mol
ta có nCa(OH)2=nCaCL2=0,1 mol
=>mCaCL2=0,1.111=11,1 gam
B) số mol của HCL là
nHCL=nCa(OH).2=0,1.2=0,2 mol
khối lượng của dung dịch HCL cần dùng
mHCL=n.M=0,2.71=14,2 gam
C)
nồng độ phần trăm là :
C/.=11,1/214,6.100/.=5/.
nNaOH = 0,2 . 2,5 = 0,5 (mol)
PTHH:
CH3COOH + NaOH -> CH3COONa + H2O
Mol: 0,5 <--- 0,5 ---> 0,5
VddCH3COOH = 0,5/2 = 0,25 (l)
VddCH3COONa = 0,25 + 0,2 = 0,45 (l)
CMddCH3COONa = 0,5/,045 = 1,11M
Ta có: \(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)
PT: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
______0,05_____0,1______0,05 (mol)
\(\Rightarrow m_{HCl}=0,1.36,5=3,65\left(g\right)\Rightarrow m_{ddHCl}=\dfrac{3,65}{7,3\%}=50\left(g\right)\)
Ta có: m dd sau pư = 4 + 50 = 54 (g)
\(\Rightarrow C\%_{CuCl_2}=\dfrac{0,05.135}{54}.100\%=12,5\%\)
a)
$CH_3COOH + NaHCO_3 \to CH_3COONa + CO_2 + H_2O$
b)
n NaHCO3 = n CH3COOH = 100.12%/60 = 0,2(mol)
m dd NaHCO3 = 0,2.84/8% = 210(gam)
c)
n CO2 = n CH3COOH = 0,2(mol)
=> V CO2 = 0,2.22,4 = 4,48(lít)
d)
m dd = m dd CH3COOH + m dd NaHCO3 - m CO2 = 100 + 210 - 0,2.44 = 301,2(gam)
C% CH3COONa = 0,2.82/301,2 .100% = 5,44%
\(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
PT: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
a, \(n_{HCl}=2n_{CuO}=0,2\left(mol\right)\Rightarrow C\%_{HCl}=\dfrac{0,2.36,5}{200}.100\%=3,65\%\)
b, \(n_{CuCl_2}=n_{CuO}=0,1\left(mol\right)\)
\(\Rightarrow C\%_{CuCl_2}=\dfrac{0,1.135}{8+200}.100\%\approx6,49\%\)