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a,\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PTHH: Zn + 2HCl ---> ZnCl2 + H2
0,1--------------->0,1------>0,1
b, => \(\left\{{}\begin{matrix}C_{M\left(ZnCl_2\right)}=\dfrac{0,1}{\dfrac{6}{1000}}=\dfrac{50}{3}M\\V_{H_2}=0,1.22,4=2,24\left(l\right)\end{matrix}\right.\)
c, \(n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: 2H2 + O2 --to--> 2H2O
LTL: \(\dfrac{0,1}{2}< 0,1\)=> O2 dư
Theo pt: \(n_{O_2}=\dfrac{1}{2}n_{H_2}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{O_2\left(dư\right)}=\left(0,1-0,05\right).32=1,6\left(g\right)\\V_{O_2\left(dư\right)}=\left(0,1-0,05\right).22,4=1,12\left(l\right)\end{matrix}\right.\)
\(n_K=\dfrac{3,9}{39}=0,1mol\)
\(2K+2H_2O\rightarrow2KOH+H_2\)
0,1 0,1 0,05 ( mol )
\(m_{KOH}=0,1.56=5,6g\)
\(V_{H_2}=0,05.22,4=1,12l\)
a, \(2K+2H_2O\rightarrow2KOH+H_2\)
b, \(n_K=\dfrac{3,9}{39}=0,1\left(mol\right)\)
Theo PT: \(n_{KOH}=n_K=0,1\left(mol\right)\Rightarrow m_{KOH}=0,1.56=5,6\left(g\right)\)
\(n_K=\dfrac{m}{M}=\dfrac{3,9}{39}=0,1\left(mol\right)\\ PTHH:2K+2H_2O->2KOH+H_2\)
tỉ lệ 2 : 2 : 2 ; 1
n(mol) 0,1---.0,1------>0,1------>0,05
\(m_{KOH}=n\cdot M=0,1\cdot\left(39+16+1\right)=5,6\left(g\right)\)
\(n_{Na}=\dfrac{2,3}{23}=0,1mol\)
\(n_{H_2O}=\dfrac{47,8}{18}=2,65mol\)
\(2Na+2H_2O\rightarrow2NaOH+H_2\)
0,1 < 2,65 ( mol )
0,1 0,1 0,05 ( mol )
\(m_{NaOH}=0,1.40=4g\)
\(m_{ddspứ}=2,3+47,8-0,05.2=50g\)
\(C\%_{NaOH}=\dfrac{4}{50}.100=8\%\)
Số mol của natri hidroxit
nNaOH = \(\dfrac{m_{NaOH}}{M_{NaOH}}=\dfrac{4}{40}=0,1\left(mol\right)\)
a) Pt : NaOH + HCl → NaCl + H2O\(|\)
1 1 1 1
0,1 0,1 0,1
b) Số mol của axit clohidric
nHCl = \(\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
Khối lượng của axit clohidric
mHCl = nHCl . MHCl
= 0,1 . 35,5
= 3,55 (g)
Số mol của natri clorua
nNaCl = \(\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
Khối lượng của natri clorua
mNaCl = nNaCl . MNaCl
= 0,1 . 58,5
= 5,85 (g)
Chúc bạn học tốt
$a\big)2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2$
$b\big)$
$n_{Al}=\dfrac{4,05}{27}=0,15(mol)$
$n_{H_2SO_4}=\dfrac{29,4}{98}=0,3(mol)$
Vì $\dfrac{n_{Al}}{2}<\frac{n_{H_2SO_4}}{3}\to H_2SO_4$ dư
$c\big)$
Theo PT: $n_{H_2}=\dfrac{3}{2}n_{Al}=0,225(mol)$
$\to V_{H_2}=0,225.22,4=5,04(l)$
a,\(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right);n_K=\dfrac{3,9}{39}=0,1\left(mol\right)\)
PTHH: 2Na + 2H2O → 2NaOH + H2
Mol: 0,2 0,1
PTHH: 2K + 2H2O → 2KOH + H2
Mol: 0,1 0,05
b, \(n_{H_2}=0,1+0,05=0,15\left(mol\right)\)
\(V_{H_2}=0,15.22,4=3,36\left(l\right)\)
c,mdd sau pứ=4,6+3,9+91,5-0,15.2=99,7 (g)
\(\%m_{NaOH}=\dfrac{0,2.40.100\%}{99,7}=8,02\%\)
\(\%m_{KOH}=\dfrac{0,1.56.100\%}{99,7}=5,62\%\)
Bài 3 :
\(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)
\(n_K=\dfrac{3,9}{39}=0,1\left(mol\right)\)
a) Pt : \(2Na+2H_2O\rightarrow2NaOH+H_2|\)
2 2 2 1
0,2 0,2 0,1
\(2K+2H_2O\rightarrow2KOH+H_2|\)
2 2 2 1
0,1 0,1 0,05
b) \(n_{H2\left(tổng\right)}=0,1+0,05=0,15\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,15.22,4=3,36\left(l\right)\)
c) \(n_{NaOH}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
⇒ \(m_{NaOH}=0,2.40=8\left(g\right)\)
\(n_{KOH}=\dfrac{0,05.2}{1}=0,1\left(mol\right)\)
⇒ \(m_{KOH}=0,1.56=5,6\left(g\right)\)
\(m_{ddspu}=8,5+91,5-\left(0,15.2\right)=99,7\left(g\right)\)
\(C_{NaOH}=\dfrac{8.100}{99,7}=8,02\)0/0
\(C_{KOH}=\dfrac{5,6.100}{99,7}=5,62\)0/0
Chúc bạn học tốt
\(n_K=\dfrac{3,9}{39}=0,1\left(mol\right)\\ a,2K+2H_2O\rightarrow2KOH+H_2\\ b,n_{KOH}=n_K=0,1\left(mol\right)\\ C\%_{ddKOH}=\dfrac{0,1.56}{150}.100\%\approx3,733\%\)
\(2K+2H_2O\rightarrow2KOH+H_2\)
\(2mol\) \(2mol\)
\(0,1mol\) \(0,1mol\)
\(n_K=\dfrac{m}{M}=\dfrac{3,9}{39}=0,1\left(mol\right)\)
\(m_{KOH}=n.M=0,1.56=5,6\left(g\right)\)
\(C\text{%}=\dfrac{m_{ct}}{m_{dd}}.100\text{%}=\dfrac{5,6}{150}.100\text{%}=\approx3,73\text{%}\)