Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(n_{Na}=\dfrac{6,9}{23}=0,3\left(mol\right)\\ 2Na+2H_2O\rightarrow2NaOH+H_2\\ n_{H_2}=\dfrac{0,3}{2}=0,15\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\\ b,m_{ddsaup.ứ}=m_{Na}+m_{H_2O}-m_{H_2}=6,9+100-0,15.2=106,6\left(g\right)\)
Fe + 2HCl \(\rightarrow FeCl_2+H_2\)
a) nFe = \(\dfrac{5,6}{56}=0,1mol\)
Theo pt nH2 = nFe = 0,1 mol
=> VH2 = 0,1.22,4 = 2,24 lít
b) Theo pt: nFeCl2 = nFe = 0,1 mol
=> mFeCl2 = 0,1.127 = 12,7g
c) Theo pt : nHCl = 2nFe = 0,2 mol
=> mHCl = 0,2.36,5 = 7,3g
\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{2,7}{27}=0,1mol\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,1 > 0,2 ( mol )
0,1 0,15 0,05 0,15 ( mol )
\(V_{H_2}=n_{H_2}.22,4=0,15.22,4=3,36l\)
\(m_{H_2SO_4\left(du\right)}=n_{H_2SO_4\left(du\right)}.M_{H_2SO_4}=\left(0,2-0,15\right).98=4,9g\)
\(m_{Al_2\left(SO_4\right)_3}=n.M=0,05.342=17,1g\)
\(n_{Al}=\dfrac{2,7}{27}=0,1mol\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,1 0,2 0 0
0,1 0,15 0,1 0,15
0 0,05 0,1 0,15
Chất dư sau phản ứng là \(H_2SO_4\) và dư 0,05mol.
\(m_{H_2SO_4dư}=0,05\cdot98=4,9g\)
\(m_{Al_2\left(SO_4\right)_3}=0,1\cdot342=34,2g\)
\(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right);n_{HCl}=\dfrac{29,2}{36,5}=0,8\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\\ a,Vì:\dfrac{0,3}{1}< \dfrac{0,8}{2}\Rightarrow HCldư\\ n_{H_2}=n_{MgCl_2}=n_{Mg}=0,3\left(mol\right)\\ n_{HCl\left(dư\right)}=0,8-0,3.2=0,2\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ b,m_{MgCl_2}=0,3.95=28,5\left(g\right)\\ m_{HCl\left(dư\right)}=0,2.36,5=7,3\left(g\right)\)
a) \(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)
\(n_{HCl}=\dfrac{29.2}{36,5}=0,8\left(mol\right)\)
PTHH : 2Mg + 2HCl -> 2MgCl + H2
Xét tỉ lệ \(\dfrac{0,3}{2}< \dfrac{0,8}{2}\)
=> HCl dư
=> \(V_{H_2}=0,075.22,4=1,68\left(l\right)\)
=> \(V_{MgCl}=0,15.22,4=3,36\left(l\right)\)
b) \(m_{H_2}=0,075.2=0,15\left(g\right)\\ m_{MgCl}=0,15.59,5=8,925\left(g\right)\)
nNa=4,6/23=0,2(mol)
PT:2Na+2H2O->2NaOH+H2
Theo Pt: nH2= 1/2n Na =1/2.0,2=0,1 (mol)
a,=>VH2=0,1.22,4=2,24(l)
b,nH2=3,36/22,4=0,15 (mol)
pt:2H2+O2->2H2O
Theo Pt: n H2O =nH2 =0,15
b,->m H2O =0,15.18=2,7 (g)
\(a,n_{Zn}=\dfrac{0,65}{65}=0,01(mol)\\ n_{HCl}=\dfrac{7,3}{36,5}=0,2(mol)\\ PTHH:Zn+2HCl\to ZnCl_2+H_2\)
Vì \(\dfrac{n_{Zn}}{1}<\dfrac{n_{HCl}}{2}\) nên \(HCl\) dư
\(\Rightarrow n_{HCl(dư)}=0,2-0,02=0,18(mol)\\ \Rightarrow m_{HCl(dư)}=0,18.36,5=6,57(g)\\ b,n_{H_2}=n_{Zn}=0,01(mol)\\ \Rightarrow V_{H_2}=0,01.22,4=0,224(l)\)
`2Na + 2H_2 O -> 2NaOH + H_2`
`0,25` `0,25` `0,125` `(mol)`
`a)V_[H_2]=0,125.22,4=2,8(l)`
`b)m_[NaOH]=0,25.40=10(g)`
Phương trình: `2Na+2H_2O -> 2NaOH + H_2` $\uparrow$
`=>n_(H_2) = 0,25 : 2 = 0,125`.
`a, V_(H_2) = 0,125 xx 22,4 = 2,8l`.
`b, m_(NaOH) = 0,25 xx 40 = 10g`.