Cho 2,4g Mg tác dụng hcl .1M. Sinh ra V lít khí h a, viết phương trình b, tính thể tích c, tính Vhcl
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\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\ pthh:Mg+2HCl\rightarrow MgCl_2+H_2\)
0,1 0,2 0,1
\(V_{H_2}=0,1.22,4=2,24l\\ m_{HCl}=\dfrac{0,2.36,5}{10}.100=73g\)
a) \(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
Theo PTHH ta có: \(n_{MgCl_2}=n_{Mg}=0,1\left(mol\right)\)
\(\Rightarrow m_{MgCl_2}=n_{MgCl_2}.M_{MgCl_2}=0,1.95=9,5\left(g\right)\)
b) Theo PTHH ta có: \(n_{H_2}=n_{Mg}=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=n_{H_2}.22,4=0,1.22,4=2,24\left(l\right)\)
c) \(2H_2+O_2\rightarrow2H_2O\)
Theo PTHH: \(n_{H_2O}=\dfrac{0,1.2}{2}=0,1\left(mol\right)\)
\(\Rightarrow m_{H_2O}=n_{H_2O}.M_{H_2O}=0,1.18=1,8\left(g\right)\)
a)mMg= 2,4/24=0,1(mol) nMgCl2=0,1.1/1=0,1 (mol) mMgCl2= 0,1.95=9,5(g) b)nH2=0,1.1/1=0,1(mol) v=n.22,4=2,24(lít
\(n_{HCl}=\dfrac{18.25}{36.5}=0.5\left(mol\right)\)
\(a.Mg+2HCl\rightarrow MgCl_2+H_2\)
\(b.\)
\(n_{Mg}=n_{H_2}=\dfrac{1}{2}\cdot n_{HCl}=\dfrac{1}{2}\cdot0.5=0.25\left(mol\right)\)
\(m_{Mg}=0.25\cdot24=6\left(g\right)\)
\(V_{H_2}=0.25\cdot22.4=5.6\left(l\right)\)
\(c.\)
\(V_{H_2\left(tt\right)}=5.6\cdot90\%=5.04\left(l\right)\)
\(n_{HCl}=0,3.1=0,3\left(mol\right)\\
pthh:Fe+2HCl\rightarrow FeCl_2+H_2\)
0,15 0,3 0,15
\(m_{Fe}=0,15.56=8,4\left(g\right)\\
V_{H_2}=0,15.22,4=3,36\left(l\right)\)
\(a)n_{Fe}=\dfrac{5,6}{56}=0,1mol\\ n_{Mg}=\dfrac{4,8}{24}=0,2mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1 0,2 0,1 0,1
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,2 0,4 0,2 0,2
\(V_{H_2}=\left(0,1+0,2\right).22,4=6,72l\\ b)V_{ddHCl}=\dfrac{0,2+0,4}{2}=0,3l\\ c)m_{muối}=0,1.127+95.0,2=31,7g\)
\(n_{Zn}=\dfrac{6,5}{65}=0,1mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1 0,2 0,1 0,1
\(V_{H_2}=0,1\cdot22,4=2,24l\)
a) $Mg + 2HCl \to MgCl_2 + H_2$
b) $n_{Mg} = \dfrac{2,4}{24} = 0,1(mol) ; n_{HCl} = 0,05.6 = 0,3(mol)$
Vì $n_{Mg} : 1 < n_{HCl} : 2$ nên $HCl$ dư
$n_{H_2} =n_{Mg} = 0,1(mol) \Rightarrow V_{H_2} = 0,1.22,4 = 2,24(lít)$
c)$n_{HCl\ pư}= 2n_{Mg} = 0,2(mol) \Rightarrow n_{HCl\ dư} = 0,3 -0,2 = 0,1(mol)$
$C_{M_{HCl}} = \dfrac{0,1}{0,05} = 2M$
$C_{M_{MgCl_2}} = \dfrac{0,1}{0,05} = 2M$
a. \(Mg+2HCl\rightarrow MgCl_2+H_2\)
b. \(n_{Mg}=\dfrac{2.4}{24}=0.1mol\)
\(mct_{HCl}=\dfrac{500\times36.5}{100}=182.5g\Rightarrow n_{HCl}=\dfrac{182.5}{36.5}=5mol\)
Ta có: \(\dfrac{0.1}{1}< \dfrac{5}{2}\Rightarrow\) HCl dư
nHCl phản ứng = 0.2 mol => nHCl dư = 5 - 0.2 = 4.8 mol
mHCl dư = \(4.8\times36.5=175.2g\)
c. \(V_{H_2}=0.1\times22.4=2.24l\)
d. mdd sau phản ứng = \(2.4+500-0.1\times2=502.2g\)
\(C\%_{MgCl_2}=\dfrac{0.1\times95\times100}{502.2}=1.89\%\)
1:
a) \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)
______0,2------>0,2------------------->0,2_____(mol)
=> \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b) \(V_{ddH_2SO_4}=\dfrac{0,2}{1}=0,2\left(l\right)\)
2:
a)
\(n_{HCl}=2.0,2=0,4\left(mol\right)\)
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
______0,2<------0,4------------------>0,2______(mol)
=> \(m_{Mg}=0,2.24=4,8\left(g\right)\)
b) \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
a) $Mg + 2HCl \to MgCl_2 + H_2$
b) $n_{H_2} = n_{Mg} = \dfrac{2,4}{24} = 0,1(mol)$
$V_{H_2} = 0,1.22,4 = 2,24(lít)$
c) $n_{HCl} = 2n_{Mg} = 0,2(mol) \Rightarrow V_{dd\ HCl} = \dfrac{0,2}{1} = 0,2(lít)$