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Zn + 2 HCl -> ZnCl2 + H2
nZn=0,2(mol) -> nH2=0,2(mol)
PTHH: Fe3O4 + 4 H2 -to-> 3 Fe + 4 H2O
nFe3O4=23,2/232=0,1(mol)
Ta có: 0,1/1 > 0,2/4
=> H2 hết, Fe3O4 dư, tính theo nH2
=> nFe=3/4. 0,2=0,15(mol)
=> mFe=0,15. 56=8,4(g)
Zn+2Hcl->ZnCl2+H2
0,2---0,4----0,2----0,2
n Zn=0,2 mol
=>VH2 =0,2.22,4=4,48l
mZncl2=0,2.136=27,2g
3H2+Fe2O3-to>2Fe+3H2O
0,2---------------------2\15
->m Fe=2\15.56=7,467g
nZn= 13/65=0,2(mol)
a) PTHH: Zn + 2 HCl -> ZnCl2 + H2
b) nH2=nZnCl2=nZn=0,2(mol)
=>V(H2,đktc)=0,2 x 22,4= 4,48(l)
c) khối lượng muối sau phản ứng chứ nhỉ?
mZnCl2=136.0,2=27,2(g)
a, Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
___0,1_________________0,1 (mol)
Ta có: \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
b, PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
Theo PT: \(n_{Fe}=\dfrac{2}{3}n_{H_2}=\dfrac{1}{15}\left(mol\right)\)
\(\Rightarrow m_{Fe}=\dfrac{1}{15}.56\approx3,73\left(g\right)\)
Bạn tham khảo nhé!
a, \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)
b, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Theo PT: \(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{15}\left(mol\right)\Rightarrow m_{Fe_3O_4}=\dfrac{1}{15}.232=\dfrac{232}{15}\left(g\right)\)
c, \(n_{H_2}=\dfrac{4}{3}n_{Fe}=\dfrac{4}{15}\left(mol\right)\Rightarrow V_{H_2}=\dfrac{4}{15}.22,4=\dfrac{448}{75}\left(l\right)\)
d, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(n_{Zn}=n_{H_2}=\dfrac{4}{15}\left(mol\right)\Rightarrow m_{Zn}=\dfrac{4}{15}.65=\dfrac{52}{3}\left(g\right)\)
\(n_{HCl}=2n_{H_2}=\dfrac{8}{15}\left(mol\right)\Rightarrow m_{HCl}=\dfrac{8}{15}.36,5=\dfrac{292}{15}\left(g\right)\)
\(nZn=\dfrac{13}{65}=0,2\left(mol\right)\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
1 1 1 1 (mol)
0,2 0,2 0,2 0,2 (mol)
\(VH_2=0,2.22,4=4,48\left(l\right)\)
\(Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)
1 3 2 3 (mol)
0,2 2/15 (mol)
\(mFe=\dfrac{2}{15}.56=7,47\left(g\right)\)
ta co pthh Zn+2HCl\(\rightarrow\)ZnCl2+H2(dknd)
theo de bai ta co nZn = \(\dfrac{13}{65}=0,2mol\)
theo pthh nH2=nZn=0,2 mol
\(\Rightarrow\)vH2= 0,2.22,4=4,488 l
ta co pthh 2 4 H2+Fe3O4\(\rightarrow\)3Fe +4 H2O(dknd)
theo cau a ta co nH2= 0,2 mol
theo de bai nFe3O4= \(\dfrac{23,2}{232}=0,1mol\)
theo pthh ta co nH2= \(\dfrac{0,2}{4}\)mol < nFe3O4= \(\dfrac{0,1}{1}mol\)
\(\Rightarrow\)nFe3O4 du tinh theo so mol cua H2
Vay khoi luong cua kim loai sat thu duoc la
mFe= (\(\dfrac{3}{4}.0,1\)).56=4,2 g
Ta có pthh 2Al +3 H2SO4 -> Al2(SO4)3 +3H2 theo đề bài ta có nAl= 2.7/27=0.1 mol , nH2SO4= 39.2/98= 0.4 mol .Theo pthh n Al=0.1/2 mol < nH2SO4= 0.4/3 mol -> nH2SO4 dư ( tính theo nAl) theo pthh nH2 = 3/2 nAl= 3/2* 0.1=0.15 mol -> vH2 = 0.15*22.4= 3.36l .Theo pthh nAl2(SO4)3=1/2* nAl=1/2*0.1= 0.05 mol -> mAl2(SO4)3= 0.05*342=17.1 g