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\(m_{H_2SO_4}=\dfrac{200.9,8}{100}=19,6\left(g\right)\)
\(n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)
PTHH :
\(X+H_2SO_4\rightarrow XSO_4+H_2\)
0,2 0,2 0,2 0,2
\(M_X=\dfrac{8}{0,2}=40\left(dvC\right)\)
-> Canxi
\(b,V_{H_2}=0,2.22,4=4,48\left(l\right)\)
\(c,m_{CaSO_4}=0,2.136=27,2\left(g\right)\)
\(m_{ddCaSO_4}=8+200-\left(0,2.2\right)=207,6\left(g\right)\)
\(C\%=\dfrac{27,2}{207,6}.100\%\approx13,1\%\)
a, Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
Theo PT: \(n_{FeSO_4}=n_{Fe}=0,1\left(mol\right)\Rightarrow m_{FeSO_4}=0,1.152=15,2\left(g\right)\)
b, \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c, Sửa đề: 500 ml → 500 (g)
Theo PT: \(n_{H_2SO_4}=n_{Fe}=0,1\left(mol\right)\Rightarrow C\%_{H_2SO_4}=\dfrac{0,1.98}{500}.100\%=1,96\%\)
a) PTHH : \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
b) \(n_{H_2SO_4}=C_MV=1,2\cdot0,5=0,6\left(mol\right)\)
PTHH : \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
0,6 0,6 0,6
\(\Rightarrow m_{FeSO_4}=n_{FeSO_4}M_{FeSO_4}=0,6\cdot152=91,2\left(g\right)\)
c) Từ câu b \(\Rightarrow n_{H_2}=0,6\left(mol\right)\)
\(\Rightarrow V_{H_2}=n_{H_2}.22,4=0,6\cdot22,4=13,44\left(l\right)\)
d) PTHH : \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
0,6 0,6
\(\Rightarrow m_{Cu}=n_{Cu}M_{Cu}=0,6\cdot64=38,4\left(g\right)\)
a)\(PTHH:Fe+H_2SO_4\xrightarrow[]{}FeSO_4+H_2\)
b)Đổi 500ml = 0,5l
Số mol của H2SO4 là:
\(C_{MH_2SO_4}=\dfrac{n_{H_2SO_4}}{V_{H_2SO_{\text{4 }}}}\Rightarrow n_{H_2SO_4}=C_{MH_2SO_4}.V_{H_2SO_4}=1,2.0,5=0,6\left(mol\right)\)
\(PTHH:Fe+H_2SO_4\xrightarrow[]{}FeSO_4+H_2\)
Tỉ lệ : 1 1 1 1 (mol)
Số mol : 0,6 0,6 0,6 0,6(mol)
Khối lượng sắt(II)sunfat thu được là:
\(m_{FeSO_4}=n_{FeSO_4}.M_{FeSO_{\text{4 }}}=0,6.152=91,2\left(g\right)\)
c) Thể tích khí H2 thoát ra là:
\(V_{H_2}=n_{H_2}.22,4=0,6.22,4=13,44\left(l\right)\)
d)\(PTHH:CuO+H_2\xrightarrow[]{t^0}Cu+H_2O\)
tỉ lệ :1 1 1 1 (mol)
số mol :0,6 0,6 0,6 0,6 (mol)
Khối lượng CuO điều chế được là:
\(m_{CuO}=n_{CuO}.M_{CuO}=0,6.80=48\left(g\right)\)
\(n_{HCl}=0,2.1=0,2\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
a, \(n_{Fe}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)
b, \(n_{FeCl_2}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\Rightarrow m_{FeCl_2}=0,1.127=12,7\left(g\right)\)
c, \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
Ta có: \(n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=1\left(mol\right)\\n_{FeCl_2}=0,5\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{ddHCl}=\dfrac{1}{2}=0,5\left(l\right)\\C_{M_{FeCl_2}}=\dfrac{0,5}{0,5}=1\left(M\right)\end{matrix}\right.\)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,3.24,79=7,437\left(l\right)\)
a, \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
b, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=n_{Fe}=0,2\left(mol\right)\Rightarrow m_{H_2SO_4}=0,2.98=19,6\left(g\right)\)
c, \(C\%_{H_2SO_4}=\dfrac{19,6}{50}.100\%=39,2\%\)
d, Theo PT: \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
PTHH: 2Na+2H2O=>2 NaOH+H2
nH2SO4=0,2mol
PTHH: 2NaOH+H2SO4=> Na2SO4+2H2O
0,4mol<-0,2mol
=> n NaOH=0,4mol
mà nNaOH=nNa=0,4mol
=> m Na =0,4.23=9,2g
nH2=1/2nNaOH=1/2.0,2=0,1mol
=> V H2=0,1.22,4=2,24ml
200 ml hay 200 l zậy bạn