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a) Mg + 2HCl --> MgCl2 + H2
b) Theo ĐLBTKL: mMg + mHCl = mMgCl2 + mH2 (1)
c)
\(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)=>m_{H_2}=0,5.2=1\left(g\right)\)
(1) => mMgCl2 = 12+36,5-1 = 47,5(g)
a: \(Mg+2HCl->MgCl_2+H_2\)
b: \(n_H=\dfrac{11.2}{22.4}=0.5\)
\(\Leftrightarrow m_H=M_H\cdot n_H=0.5\cdot2=1\left(g\right)\)
\(a.Mg+2HCl\rightarrow MgCl_2+H_2\\ b.m_{H_2}=\dfrac{11,2}{22,4}.2=1\left(g\right)\\ c.BTKL:m_{Mg}+m_{HCl}=m_{MgCl_2}+m_{H_2}\\ \Rightarrow m_{MgCl_2}=12+36,5-1=47,5\left(g\right)\)
a) Mg + 2HCl --> MgCl2 + H2
b) Theo ĐLBTKL: mMg + mHCl = mMgCl2 + mH2 (1)
c) \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
=> \(m_{H_2}=2.0,5=1\left(g\right)\)
Theo ĐLBTKL => mMgCl2 =mMg + mHCl - mH2
= 12+36,5-1 = 47,5(g)
nMg = 3,6 : 24 = 0,15 (mol)
pthh : Mg + 2HCl --> MgCl2 + H2
0,15-----------> 0,15 --->0,15 (mol)
mMgCl2 = 0,15 . 95 = 14,25 (mol)
VH2 (đkc)= 0,15. 24,79 = 3,718(l)
\(n_{Mg}=\dfrac{3,6}{24}=0,15mol\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,15 0,3 0,15 0,15
\(m_{MgCl_2}=0,15\cdot95=14,25g\)
\(V_{H_2}=0,15\cdot22,4=3,36l\)
200ml = 0,2l
\(n_{HCl}=0,5.0,2=0,1\left(mol\right)\)
a) Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)
1 2 1 1
0,05 0,1 0,1 0,05
b) \(n_{Mg}=\dfrac{0,1.2}{1}=0,05\left(mol\right)\)
⇒ \(m_{Mg}=0,05.24=1,2\left(g\right)\)
c) \(n_{H2}=\dfrac{0,1.1}{2}=0,05\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,05.22,4=1,12\left(l\right)\)
d) \(n_{MgCl2}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)
\(C_{M_{MgCl2}}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)
Chúc bạn học tốt
a) Zn + 2HCl --> ZnCl2 + H2
b) Theo ĐLBTKL: mZn + mHCl = mZnCl2 + mH2
=> mH2 = 5,2 + 5,84 - 10,88 = 0,16 (g)
c) \(n_{H_2}=\dfrac{0,16}{2}=0,08\left(mol\right)\)
=> VH2 = 0,08.22,4 = 1,792(l)
\(a) Mg + 2HCl \to MgCl_2 + H_2\\ b) n_{MgCl_2} = n_{Mg} = \dfrac{0,24}{24} = 0,01(mol)\\ m_{MgCl_2} = 0,01.95 = 0,95(gam)\\ c) n_{H_2} = n_{Mg} = 0,01(mol) \Rightarrow V_{H_2} = 0,01.22,4 = 0,224(lít)\)
\(n_{Mg}=\dfrac{3,12}{24}=0,13\left(mol\right)\)
\(n_{H_2}=\dfrac{2,2311}{25,79}=0,09\left(mol\right)\)
PTHH :
\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
0,09 0,18 0,09 0,09
\(\dfrac{0,13}{1}>\dfrac{0,09}{1}\) --> Mg dư
\(m_{MgCl_2}=0,09.95=8,55\left(g\right)\)
PTHH :
\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
0,13 0,26 0,13 0,13
\(V_{H_2\left(lt\right)}=0,13.24,79=3,2227\left(l\right)\)
\(H=\dfrac{2,2311}{3,2227}.100\%\approx69,23\%\)
a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)
b, \(m_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)
Theo PT: \(n_{MgCl_2}=n_{Mg}=0,3\left(mol\right)\Rightarrow m_{MgCl_2}=0,3.95=28,5\left(g\right)\)
c, \(n_{HCl}=2n_{Mg}=0,6\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,2}=3\left(M\right)\)
\(a,Mg+2HCl\to MgCl_2+H_2\\ b,m_{H_2}=\dfrac{11,2}{22,4}.2=1(g)\\ c,BTKL:m_{MgCl_2}=m_{Mg}+m_{HCl}-m_{H_2}=12+36,5-1=47,5(g)\)