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\(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)
Pt : \(4Na+O_2\underrightarrow{t^o}2Na_2O|\)
4 1 2
0,2 0,1
\(Na_2O+2HCl\rightarrow2NaCl+H_2O|\)
1 2 2 1
0,2 0,4
b) \(n_{Na2O}=\dfrac{0,2.2}{4}=0,1\left(mol\right)\)
⇒ \(m_{Na2O}=0,1.62=6,2\left(g\right)\)
c) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)
\(V_{ddHCl}=\dfrac{0,4}{1}=0,4\left(l\right)\) = 400 (ml)
Chúc bạn học tốt
\(n_P=\dfrac{m}{M}=\dfrac{12,4}{31}=0,4\left(mol\right)\\ PTHH:4P+5O_2-^{t^o}>2P_2O_5\)
Tỉ lệ 4 : 5 : 2
n(mol) 0,4--->0,5----->0,2
\(V_{O_2\left(dktc\right)}=n\cdot22,4=0,5\cdot22,4=11,2\left(l\right)\\ V_{kk}=11,2:\dfrac{1}{5}=56\left(l\right)\)
\(m_{P_2O_5}=n\cdot M=0,2\cdot142=28,4\left(g\right)\)
\(PTHH:P_2O_5+3H_2O->2H_3PO_4\)
tỉ lệ 1 : 3 : 2
n(mol) 0,2----->0,6--------->0,4
\(m_{H_3PO_4}=n\cdot M=0,4\cdot98=39,2\left(g\right)\)
\(C\%=\dfrac{m_{ct}}{m_{dd}}\cdot100\%=\dfrac{39,2}{200}\cdot100\%=19,6\%\)
Gọi n H2O = a(mol)
n H2 = 3,36/22,4 = 0,15(mol)
$MO + 2HCl \to MCl_2 + H_2O$
$M + 2HCl \to MCl_2 + H_2$
n HCl = 2n H2O + 2n H2 = 2a + 0,3(mol)
Bảo toàn khối lượng :
9,6 + (2a + 0,3)36,5 = 28,5 + 18a + 0,15.2
=> a = 0,15(mol)
n MO = n H2O = 0,15(mol)
n M = n H2 = 0,15(mol)
=> 0,15(M + 16) + 0,15M = 9,6
=> M = 24(Mgaie)
n Mg= n MgO + n Mg = 0,3(mol)
=> a = 0,3.24 = 7,2 gam
a) \(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)
PTHH: 2Na + 2H2O ---> 2NaOH + H2 => ddA là NaOH
0,2----------------->0,2------>0,1
b) \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c) \(C_{M\left(NaOH\right)}=\dfrac{0,2}{0,4}=0,5M\)
\(n_K=\dfrac{7,8}{39}=0,2\left(mol\right)\\
pthh:2K+2H_2O\rightarrow2KOH+H_2\uparrow\)
QT chuyển xanh
\(pthh:2K+2H_2O\rightarrow2KOH+H_2\)
0,2 0,2 0,1
\(V_{H_2}=0,1.22,4=2,24\left(L\right)\\
m_{KOH}=0,2.56=11,2\left(g\right)\)
\(pthh:Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)
0,1 0,075
=> \(m_{Fe}=\left(0,075.56\right).80\%=3,36g\)
\(n_{Cl_2\left(pư\right)}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: 2M + nCl2 --to--> 2MCln
0,3------>\(\dfrac{0,6}{n}\)
MCln + nNaOH + M(OH)n + nNaCl
\(\dfrac{0,6}{n}\)------------->\(\dfrac{0,6}{n}\)
=> \(\dfrac{0,6}{n}\left(M_M+17n\right)=21,4\)
=> \(M_M=\dfrac{56}{3}n\left(g/mol\right)\)
Xét n = 3 thỏa mãn => MM = 56 (g/mol)
=> M là Fe
a)
$2Al +3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
b)
$n_{H_2} = n_{H_2SO_4} = \dfrac{300.9,8\%}{98} = 0,3(mol)$
$V_{H_2} = 0,3.22,4 = 6,72(lít)$
c)
$n_{Al_2(SO_4)_3} = \dfrac{1}{3}n_{H_2SO_4} = 0,1(mol)$
$m_{Al_2(SO_4)_3} = 0,1.342 = 34,2(gam)$
d)
$n_{Al} = \dfrac{2}{3}n_{H_2SO_4} = 0,2(mol)$
$m_{dd} = 0,2.27 + 300 - 0,3.2 = 304,8(gam)$
$C\%_{Al_2(SO_4)_3} = \dfrac{34,2}{304,8}.100\% = 11,22\%$
nH2SO4=0,3(mol)
PTHH: 2Al + 3 H2SO4 -> Al2(SO4)3 + 3 H2
a) 0,2_______0,3______0,1______0,3(mol)
b) V(H2,đktc)=0,3.22,4=6,72(l)
c) a=mAl=0,2.27=5,4(g)
=>a=5,4(g)
d) mAl2(SO4)3=342.0,1=34,2(g)
e) mddAl2(SO4)3= 5,4+ 300 - 0,3.2= 304,8(g)
=>C%ddAl2(SO4)3= (34,2/304,8).100=11,22%
a, \(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)
PTHH: 4Na + O2 ---to→ 2Na2O
Mol: 0,2 0,1
PTHH: Na2O + 2HCl → 2NaCl + H2O
Mol: 0,1 0,2
b, \(m_{Na_2O}=0,1.62=6,2\left(g\right)\)
c, \(V=V_{ddHCl}=\dfrac{0,2}{1}=0,2\left(l\right)=200\left(ml\right)\)