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\(a.Mg+2HCl\rightarrow MgCl_2+H_2\\ b.n_{Mg}=\dfrac{4,8}{24}=0,2mol\\ Mg+2HCl\rightarrow MgCl_2+H_2\)
0,2 0,4 0,2 0,2
\(V_{H_2}=0,2.24,79=4,958l\\ c.m_{MgCl_2}=0,2.95=19g\\ d.C_{M_{HCl}}=\dfrac{0,4}{0,3}=\dfrac{4}{3}M\)
Ta có: \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
Theo PT: \(n_{MgCl_2}=n_{H_2}=n_{Mg}=0,1\left(mol\right)\)
a, \(m_{MgCl_2}=0,1.95=9,5\left(g\right)\)
b, \(V_{H_2}=0,1.24,79=2,479\left(l\right)\)
c, \(n_{HCl}=2n_{Mg}=0,2\left(mol\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{0,2.36,5}{3,65\%}=200\left(g\right)\)
PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
Làm gộp các phần còn lại
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=0,1mol\\n_{H_2SO_4}=n_{H_2}=0,3mol\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\m_{Al_2\left(SO_4\right)_3}=0,1\cdot342=34,2\left(g\right)\\m_{H_2SO_4}=0,3\cdot98=29,4\left(g\right)\end{matrix}\right.\)
\(n_{Zn}=\dfrac{3,9}{65}=0,06mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
0,06 0,12 0,06 0,06
\(V_{H_2}=0,06\cdot22,4=1,344l\)
\(d_{H_2}\)/CO2=\(\dfrac{M_{H_2}}{M_{CO_2}}=\dfrac{2}{44}=\dfrac{1}{22}\)
\(m_{HCl}=0,12\cdot36,5=4,38g\)
\(m_{ZnCl_2}=0,06\cdot136=8,16g\)
a) Zn + 2HCl ---> ZnCl2 + H2
b) nZn = 3,9:65= 0,06 ( mol)
theo pt , nH2 =nZn= 0,06 (mol)
=> VH2(ĐKTC) = 0,06.22,4=1,344(l)
H2/CO2 = MH2/MCO2 =2/44=1/22
c) theo pt nHCl = 2nZn = 2.0,06=0,12(mol)
=> mHCl= 0,12 . 36,5=4,38(g)
d) theo pt , nZnCl2= nZn = 0,06(mol)
=> m ZnCl2 = 0,06.136=8,16 (g)
a)
\(PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
b)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
Theo PT:\(n_{H_2}=n_{Zn}=0,2mol\)
\(\Rightarrow V_{H_2}=0,2.24,79=4,958l\)
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Zn}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2.24,79=4,958\left(l\right)\)
a) PTHH: Mg + 2HCl -> MgCl2 + H2
b) Theo ĐLBTKL
\(m_{Mg}+m_{HCl}=m_{MgCl_2}+m_{H_2}\\ =>9,6+300=m_{MgCl_2}+0,8\)
c) Khối lượng MgCl2 thu dc là
\(m_{MgCl_2}=309,6-0,8=308,8\left(g\right)\)
a, PTHH : \(Mg+2HCl -> MgCl_2+H_2\)
b/ Công thức ĐLBTLKL: \(m_{Mg}+m_{HCl}=m_{MgCl_2}+m_{H_2}\)
c/ Áp dụng ĐLBTKL, ta có: \(m_{Mg}+m_{HCl}=m_{MgCl_2}+m_{H_2}\)
\(=> m_{MgCl_2}=(9,6+300)-0,8=308,8(g)\)
Vậy khối lượng \(MgCl_2\) thu được là \(308,8 g \)
a,Mg+HCL->MgCL2+H2
b,
nMg=m/M=4,8/24=0,2(mol)
PTHH:Mg+2HCl->MgCl2+H2
theo PT:1mol 2mol 1mol 1mol
theo bài ra: 0,2mol x y z
mMgCL2=n*M=0,2*59,5=11,9(g)
nH2=0,2*1/1=0,2(mol)
VH2=n*24,79=0,2*24,79=4,958(L)