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\(n_{Na2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)
Pt : \(Na_2O+2HCl\rightarrow2NaCl+H_2O|\)
1 2 2 1
0,25 0,5 0,5
a) \(n_{NaCl}=\dfrac{0,25.2}{1}=0,5\left(mol\right)\)
⇒ \(m_{NaCl}=0,5.58,5=29,25\left(g\right)\)
b) \(n_{HCl}=\dfrac{0,25.2}{1}=0,5\left(mol\right)\)
\(C_{M_{ddHCl}}=\dfrac{0,5}{0,5}=1\left(M\right)\)
Chúc bạn học tốt
\(n_{Na_2SO_3}=\dfrac{12,6}{126}=0,1mol\)
\(Na_2SO_3+H_2SO_4\rightarrow Na_2SO_4+SO_2+H_2O\)
a) \(n_{SO_2}=n_{Na_2SO_3}=0,1mol\) \(\Rightarrow V=2,24l\)
b) \(n_{H_2SO_4}=n_{Na_2SO_3}=0,1mol\) \(\Rightarrow C_M=\dfrac{0,1}{0,2}=0,5M\)
c) \(m_{Na_2SO_4}=0,1\cdot142=14,2g\)
\(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
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) \(n_{CuCl2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)
⇒ \(m_{CuCl2}=0,1.135=13,5\left(g\right)\)
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\(n_{H_2SO_4}=0,05\cdot3=0,15mol\)
a) \(H_2SO_4+BaCl_2\rightarrow BaSO_4\downarrow+2HCl\)
0,15 0,15 0,15 0,3
\(m_{ctBaCl_2}=0,15\cdot208=31,5\left(g\right)\)
\(m_{BaCl_2thamgia}=\dfrac{31,5}{20\%}\cdot100\%=157,5\left(g\right)\)
b) \(m_{BaSO_4}=0,15\cdot233=34,95\left(g\right)\)
c) \(Ca\left(OH\right)_2+H_2SO_4\rightarrow CaSO_4+2H_2O\)
0,15 0,15
\(m_{ctCa\left(OH\right)_2}=0,15\cdot74=11,1\left(g\right)\)
\(m_{ddCa\left(OH\right)_2}=\dfrac{11,1}{25\%}\cdot100\%=44,4\left(g\right)\)
\(\Rightarrow V_{Ca\left(OH\right)_2}=\dfrac{44,4}{1,15}=38,6\left(ml\right)\)
a) \(n_{Fe}=\dfrac{1,12}{56}=0,02\left(mol\right)\)
PTHH: Fe + 2HCl -->FeCl2 + H2
_____0,02->0,04--->0,02--->0,02
=> VH2 = 0,02.22,4 = 0,448(l)
b) mFeCl2 = 0,02.127 = 2,54(g)
c) \(C_{M\left(HCl\right)}=\dfrac{0,04}{0,2}=0,2M\)
Fe + 2HCl → FeCl2 + H2
1 2 1 1
0,02 0,04 0,02 0,02
nFe=\(\dfrac{1,12}{56}\)= 0,02(mol)
a). nH2=\(\dfrac{0,02.1}{1}\)= 0,02(mol)
→VH2= n . 22,4 = 0,02 . 22,4 = 0,448(l)
b). nFeCl2= \(\dfrac{0,02.1}{1}\)= 0,02(mol)
→mFeCl2= n . M = 0,02 . 127 = 2,54(g)
c). 200ml = 0,2l
nHCl= \(\dfrac{0,02.2}{1}\)=0,04(mol)
→CM= \(\dfrac{n}{V}\)= \(\dfrac{0,04}{0,2}\)= 0,2M
\(n_{CuO}=\dfrac{1,6}{80}=0,02\left(mol\right)\)
\(m_{H_2SO_4}=100.20\%=20\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{20}{98}=\dfrac{10}{49}\left(mol\right)\)
PTHH: CuO + H2SO4 → CuSO4 + H2O
Mol: 0,02 0,02 0,02
Ta có: \(\dfrac{0,02}{1}< \dfrac{\dfrac{10}{49}}{1}\) ⇒ CuO hết, H2SO4 dư
mdd sau pứ = 1,6 + 100 = 101,6 (g)
\(C\%_{ddCuSO_4}=\dfrac{0,02.160.100\%}{101,6}=3,15\%\)
\(C\%_{ddH_2SO_4}=\dfrac{\left(\dfrac{10}{49}-0,02\right).98.100\%}{101,6}=17,76\%\)
\(a.HCl+NaOH\rightarrow NaCl+H_2O\)
PỨ trung hoà
\(b,n_{NaOH}=0,1.1=0,1mol\\ n_{NaCl}=n_{NaOH}=n_{HCl}0,1mol\\ m=m_{HCl}=0,1.36,5=3,65g\\ c,m_{NaCl}=0,1.58,5=5,85g\\ d,n_{HCl}=\dfrac{73.10}{100.36,5}=0,2mol\\ \Rightarrow\dfrac{0,1}{1}< \dfrac{0,2}{1}\Rightarrow HCl.dư\\ n_{HCl,pứ}=n_{NaOH}=0,1mol\\ m_{HCl,dư}=\left(0,2-0,1\right).36,5=3,65g\)
a)
$Fe + CuSO_4 \to FeSO_4 + Cu$
Theo PTHH : $n_{Cu} = n_{CuSO_4} = 0,3.1 = 0,3(mol)$
$m_{Cu} = 0,3.64 = 19,2(gam)$
b) $n_{FeSO_4} = n_{CuSO_4} = 0,3(mol)$
$\Rightarrow m_{FeSO_4} = 0,3.152 = 45,6(gam)$
c) $FeSO_4 + 2NaOH \to Fe(OH)_2 + Na_2SO_4$
$n_{Fe(OH)_2} = n_{FeSO_4} = 0,3(mol)$
$m_{Fe(OH)_2} = 0,3.90 = 27(gam)$
Chắc là 1,6g CuO
\(n_{CuO}=\dfrac{1,6}{160}=0,01\left(mol\right)\)
a) Pt : \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O|\)
1 1 1 1
0,01 0,01 0,01
b) \(n_{H2SO4}=\dfrac{0,01.1}{1}=0,01\left(mol\right)\)
⇒ \(m_{H2SO4}=0,01.98=0,98\left(g\right)\)
c) \(n_{CuSO4}=\dfrac{0,0.1}{1}=0,01\left(mol\right)\)
⇒ \(m_{CuSO4}=0,01.160=1,6\left(g\right)\)
Chúc bạn học tốt