Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH: 2R + 2H2O --> 2ROH + H2
0,4<------------------0,2
=> \(M_R=\dfrac{9,2}{0,4}=23\left(g/mol\right)\)
=> R là Na
Bài 1:
\(PTHH:2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ n_{NaOH}=\dfrac{6}{40}=0,15\left(mol\right)\\ \Rightarrow n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}\cdot0,15=0,075\left(mol\right)\\ \Rightarrow m=m_{H_2SO_4}=0,075\cdot98=7,35\left(g\right)\\ n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ \Rightarrow n_{H_2}=0,2\left(mol\right)\\ \Rightarrow V=V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\)
\(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\\ n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ PTHH:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ LTL:0,2>0,15\Rightarrow CuO.dư\\ Theo.pt:n_{CuO\left(pư\right)}=n_{Cu}=n_{H_2}=0,15\left(mol\right)\\ \Rightarrow m_{chất.rắn}=\left(0,2-0,15\right).80+64.0,15=13,6\left(g\right)\)
a)
\(CuO + H_2 \xrightarrow{t^o} Cu + H_2O\\ Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O\\ Fe_3O_4 + 4H_2 \xrightarrow{t^o} 3Fe + 4H_2O\\ FeO + H_2 \xrightarrow{t^o} Fe + H_2O\)
b)
\(n_{H_2} = n_{H_2O} = \dfrac{14,4}{18} = 0,8(mol)\\ \Rightarrow m = m_X + m_{H_2} - m_{H_2O} = 64 + 0,8.2 - 14,4 = 51,2(gam)\)
Câu 2 :
\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)
\(CuO+H_2\underrightarrow{^{t^0}}Cu+H_2O\)
\(K+H_2O\rightarrow KOH+\dfrac{1}{2}H_2\)
\(4P+5O_2\underrightarrow{^{ }t^0}2P_2O_5\)
\(3Fe+2O_2\underrightarrow{^{t^0}}Fe_3O_4\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
Câu 3 :
\(n_{H_2}=\dfrac{2.7}{27}=0.1\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(0.1...........................0.05.......0.15\)
\(V_{H_2}=0.15\cdot22.4=3.36\left(l\right)\)
\(m_{Al_2\left(SO_4\right)_3}=0.05\cdot342=17.1\left(g\right)\)
\(m_{\text{dung dịch sau phản ứng}}=2.7+200-0.15\cdot2=202.4\left(g\right)\)
\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{17.1}{202.4}\cdot100\%=8.44\%\)
Gọi: n là hóa trị của kim loại A
Đặt: nA= x mol
nA2On= y mol
mhh= Ax + y(2A + 16n) = 30.5g
<=> Ax + 2Ay + 16yn= 30.5 (1)
<=> A(x + 2y) + 16yn= 30.5 (2)
2A + 2nH2O --> 2A(OH)n + nH2
x______________x________xn/2
A2On + nH2O --> 2A(OH)n
y_______________2y
dd B : A(OH)n
nH2= 3.36/22.4=0.15 mol
<=> xn/2= 0.15
<=> xn= 0.3 <=> x= 0.3/n
mA(OH)n= (x + 2y)(A+17n)= 39.2g
<=> Ax + 17nx + 2Ay + 34yn = 39.2
<=> Ax + 17*0.3 + 2Ay + 34yn= 39.2
<=> Ax + 2Ay + 34yn = 34.1 (3)
Lấy (3) - (1):
18yn= 3.6
<=> yn= 0.2 => y= 0.2/n
Thay x, y vào (2):
A ( 0.3/n + 0.2/n*2) + 16*0.2=30.5
<=> 0.7/n * A= 27.3
<=> A= 39n
Biện luận:
n=1 => A= 39 (K)
n=2 => A= 75 (l)
n=3 => A= 117 (l)
Vậy: A là kali (K)
`Zn + 2HCl -> ZnCl_2 + H_2↑`
`0,2` `0,4` `0,2` `0,2` `(mol)`
`a) n_[Zn] = 13 / 65 = 0,2 (mol)`
`-> V_[H_2] = 0,2 . 22,4 = 4,48 (l)`
_________________________________________
`b) m_[dd HCl] = [ 0,4 . 36,5 ] / [7,3] . 100 = 200 (g)`
_________________________________________
`c) C%_[ZnCl_2] = [ 0,2 . 136 ] / [ 200 + 13 - 0,2 . 2 ] . 100 ~~ 12,79%`
Zn+2HCl→ZnCl2+H2↑Zn+2HCl→ZnCl2+H2↑
0,20,2 0,40,4 0,20,2 0,20,2 (mol)(mol)
a)nZn=1365=0,2(mol)a)nZn=1365=0,2(mol)
→VH2=0,2.22,4=4,48(l)→VH2=0,2.22,4=4,48(l)
_________________________________________
b)mddHCl=0,4.36,57,3.100=200(g)b)mddHCl=0,4.36,57,3.100=200(g)
_________________________________________
c)C%ZnCl2=0,2.136200+13−0,2.2.100≈12,79%.
nH2 = 3,36/22,4 = 0,15 (mol)
PTHH: Ba + 2H2O -> Ba(OH)2 + H2
nBa = nH2 = 0,15 (mol)
mBa = 0,15 . 137 = 20,55 (g)
lag phím à