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\(2Na+2H_2O\rightarrow2NaOH+H_2\\2K+2H_2O\rightarrow2KOH+H_2\\ n_{Na}=0,1\left(mol\right);n_K=0,05\left(mol\right)\\ n_{H_2}=\dfrac{1}{2}n_{Na}+\dfrac{1}{2}n_K=0,05+0,025=0,075\left(mol\right)\\ \Rightarrow V_{H_2}=0,075.22,4=1,68\left(l\right)\)
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
\(a,PTHH:2Na+2H_2O\rightarrow2NaOH+H_2\\ 2K+2H_2O\rightarrow2KOH+H_2\\ n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right);n_K=\dfrac{3,9}{39}=0,1\left(mol\right)\\ b,n_{H_2\left(tổng\right)}=\dfrac{1}{2}.\left(n_{Na}+n_K\right)=\dfrac{0,2+0,1}{2}=0,15\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)
\(n_{Na}=\dfrac{4.6}{23}=0.2\left(mol\right)\)
\(n_K=\dfrac{3.9}{39}=0.1\left(mol\right)\)
\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)
\(K+H_2O\rightarrow KOH+\dfrac{1}{2}H_2\)
\(n_{H_2}=0.1+0.05=0.15\left(mol\right)\)
\(V_{H_2}=0.15\cdot22.4=3.36\left(l\right)\)
a) nNa=4,6/23=0,2(mol)
nK=3,9/39=0,1(mol)
PTHH: 2 Na + 2 H2O -> 2 NaOH + H2
0,2____________0,2______0,2__0,1(mol)
2 K + 2 H2O -> 2 KOH + H2
0,1____0,1______0,1___0,05(mol)
b) V(H2,đktc)=(0,05+0,1).22,4=3,36(l)
Bài 13: nNa= 0,2 mol ; nK= 0,1 mol
2Na + 2H2O → 2NaOH + H2↑
0,2 mol 0,2 mol 0,1 mol
2K + 2H2O → 2KOH + H2↑
0,1 mol 0,1 mol 0,05 mol
a) tổng số mol khí H2 là: nH2= 0,1 + 0,05 = 0,15 mol
→VH2= 0,15 x 22,4 = 3,36 (l)
b) mNaOH= 0,2 x 40= 8 (g) ; mKOH= 0,1 x 56= 5,6 (g)
mdung dịch= mNa + mK + mH2O - mH2 = 4,6 + 3,9 + 91,5 - 0,15x2 = 99,7 (g)
→C%NaOH= 8/99,7 x100%= 8,02%
→C%KOH= 5,6/99,7 x100%= 5,62%
a) \(2Na+2H_2O\rightarrow2NaOH+H_2\left(1\right)\)
\(2K+2H_2O\rightarrow2KOH+H_2\left(2\right)\)
b) \(n_{Na}=\frac{4,6}{23}=0,2\left(mol\right)\)
Theo PTHH (1): \(n_{Na}:n_{H_2}=2:1\)
\(\Rightarrow n_{H_2\left(1\right)}=n_{Na}.\frac{1}{2}=0,2.\frac{1}{2}=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2\left(1\right)}=0,1.22,4=2,24\left(l\right)\)
\(n_K=\frac{3,9}{39}=0,1\left(mol\right)\)
Theo PTHH (2): \(n_K:n_{H_2}=2:1\)
\(\Rightarrow n_{H_2\left(2\right)}=n_K.\frac{1}{2}=0,1.\frac{1}{2}=0,05\left(mol\right)\)
\(\Rightarrow V_{H_2\left(2\right)}=0,05.22,4=1,12\left(l\right)\)
\(\Rightarrow V_{h^2}=2,24+1,12=3,36\left(l\right)\)
c) Dung dịch thu được sau phản ứng làm giấy quỳ tím chuyển đổi thành màu xanh vì nó là dung dịch bazơ.
`2Na+2H_2O->2NaOH+H_2`
x-----------------------------`1/2`x mol
`2K+2H_2O->2KOH+H_2`
y---------------------------`1/2` y mol
`n_(H_2)=(6,72)/(22,4)=0,3 mol`
Ta có phương trình :
\(\left\{{}\begin{matrix}23x+39y=9,3\\\dfrac{1}{2}x+\dfrac{1}{2}y=0,3\end{matrix}\right.\)
-> nghiệm vô lí
`#YBTran~`
\(n_K=\dfrac{m}{M}=\dfrac{7,8}{39}=0,2\left(mol\right)\)
\(a,PTHH:4K+O_2\rightarrow2K_2O\)
\(0,2:0,05:0,1\left(mol\right)\)
\(K_2O+H_2O\rightarrow2KOH\)
\(0,1:0,1:0,2\left(mol\right)\)
\(b,V_{O_2}=n.22,4=0,05.22,4=1,12\left(l\right)\)
\(c,m_{KOH}=n.M=0,2.\left(39+16+1\right)=0,2.56=11,2\left(g\right)\)
nK=mM=7,839=0,2(mol)��=��=7,839=0,2(���)
a,PTHH:4K+O2→2K2O�,����:4�+�2→2�2�
0,2:0,05:0,1(mol)0,2:0,05:0,1(���)
K2O+H2O→2KOH�2�+�2�→2���
0,1:0,1:0,2(mol)0,1:0,1:0,2(���)
b,VO2=n.22,4=0,05.22,4=1,12(l)�,��2=�.22,4=0,05.22,4=1,12(�)
c,mKOH=n.M=0,2.(39+16+1)=0,2.56=11,2(g)�,����=�.�=0,2.(39+16+1)=0,2.56=11,2(�)
nNa=2,3/23=0,1(mol); nK=7,8/39=0,2(mol)
PTHH: Na + H2O -> NaOH + 1/2 H2
0,1_______________________0,05(mol)
K + H2O -> KOH + 1/2 H2
0,2____________0,1(mol)
=> nH2(tổng)=0,05+0,1=0,15(mol)
=>V(H2,đktc)=0,15 x 22,4= 3,36(l)