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\(m_{NaOH\left(bđ\right)}=\dfrac{90,7.8}{100}=7,256\left(g\right)\)
\(n_{Na_2O}=\dfrac{a}{62}\left(mol\right)\)
PTHH: Na2O + H2O --> 2NaOH
\(\dfrac{a}{62}\)------------->\(\dfrac{a}{31}\)
=> \(m_{NaOH\left(sau.pư\right)}=\dfrac{a}{31}.40+7,256\left(g\right)\)
mdd sau pư = a + 90,7 (g)
=> \(C\%_{dd.sau.pư}=\dfrac{\dfrac{40}{31}a+7,256}{a+90,7}.100\%=12\%\)
=> a = 3,1 (g)
Theo đề bài ta có :
nNa=\(\dfrac{a}{23}mol\)
Ta có PTHH : 1
\(2Na+2H2O\rightarrow2NaOH+H2\)
a/23.........a/23.........a/46
X=\(\dfrac{mct\left(mNaOH\right)}{m\text{dd}NaOH}.100\%=\dfrac{\left(\dfrac{a}{23}\right).40}{a+p-\left(\dfrac{a}{46}\right)}=\dfrac{40a}{23a+23p}\)(1)
Ta có : nNa2O=\(\dfrac{b}{62}mol\)
PTHH 2 :
\(Na2O+H2O\rightarrow2NaOH\: \)
b/62mol................2.(b/62)mol
=> X=\(\dfrac{mct}{m\text{dd}}.100\%=\dfrac{\left(2.\left(\dfrac{b}{62}\right)\right).40}{b+p}=\dfrac{40b}{124b+124p}\left(2\right)\)
Ta cho (1)=(2)
ta có biểu thức :
\(\dfrac{40a}{23a+23p}=\dfrac{40b}{124b+124p}\)
Bạn tự rút gọn biểu thức nhé!
CHO MÌNH CHỮA LẠI 1 CHÚT nãy ghi nhầm =.=
Theo đề bài ta có :
nNa=\(\dfrac{a}{23}mol\)
PTHH 1 :
\(2Na+2H2O\rightarrow2NaOH+H2\uparrow\)
a/23mol..................2(a/23)mol....1/2(a/23)mol
=> X=\(\dfrac{mct}{m\text{dd}}.100\%=\dfrac{40.2.\left(\dfrac{a}{23}\right)}{a+p-2.\dfrac{1}{2}\left(\dfrac{a}{23}\right)}\)=\(\dfrac{80a}{23a+23p}\)(1)
Theo đề bài ta có :
nNa2O=\(\dfrac{b}{62}mol\)
PTHH 2 :
\(Na2O+H2O\rightarrow2NaOH\)
b/62mol.................2(b/62)mol
=> X= \(\dfrac{mct}{m\text{đ}}.100\%=\dfrac{40.2\left(\dfrac{b}{62}\right)}{b+p}=\dfrac{80b}{62b+62p}\left(2\right)\)
Cho (1)=(2)
Ta có biểu thức : \(\dfrac{80a}{23a+23p}\)= \(\dfrac{80b}{62b+62p}\)
a, \(Na_2O+H_2O\rightarrow2NaOH\)
Ta có: \(n_{Na_2O}=\dfrac{3,1}{62}=0,05\left(mol\right)\)
Theo PT: \(n_{NaOH}=2n_{Na_2O}=0,1\left(mol\right)\)
m dd sau pư = 3,1 + 50 = 53,1 (g)
\(\Rightarrow C\%_{NaOH}=\dfrac{0,1.40}{53,1}.100\%\approx7,53\%\)
b, \(2Na+2H_2O\rightarrow2NaOH+H_2\)
Ta có: \(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{NaOH}=n_{Na}=0,2\left(mol\right)\\n_{H_2}=\dfrac{1}{2}n_{Na}=0,1\left(mol\right)\end{matrix}\right.\)
Ta có: m dd sau pư = 4,6 + 95,6 - 0,1.2 = 100 (g)
\(\Rightarrow C\%_{NaOH}=\dfrac{0,2.40}{100}.100\%=8\%\)
Gọi số mol NaOH là a (mol)
\(n_{Ba\left(OH\right)_2}=\dfrac{20,52}{171}=0,12\left(mol\right)\); \(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
Ta có sơ đồ:
\(21,9\left(g\right)X\left\{{}\begin{matrix}Na\\Ba\\Na_2O\\BaO\end{matrix}\right.+H_2O\rightarrow\left\{{}\begin{matrix}Ba\left(OH\right)_2:0,12\left(mol\right)\\NaOH:a\left(mol\right)\end{matrix}\right.+H_2:0,05\left(mol\right)\)
Bảo toàn H: \(n_{H_2O}=\dfrac{0,12.2+a+0,05.2}{2}=0,17+0,5a\left(mol\right)\)
Bảo toàn khối lượng:
\(m_X+m_{H_2O}=m_{Ba\left(OH\right)_2}+m_{NaOH}+m_{H_2O}\)
=> \(21,9+18\left(0,17+0,5a\right)=20,52+40a+0,05.2\)
=> a = 0,14 (mol)
=> m = 0,14.40 = 5,6 (g)
Na2O=0,5 mol
Na2O+H2O->2NaOH
0,5-----------------1 mol
ta có m NaOH=1.40+40=80g
=>C%=\(\dfrac{80}{431}100=18,561\%\)
\(C\%_{NaOH}=\dfrac{a}{a+40}=20\%\\ \Leftrightarrow a=10\left(g\right)\)
Ở 200C, 100 (g) nước hòa tan 31.6 (g) KNO3 tạo thành dung dịch bão hòa.
Ở 200C, 106.5 (g) nước hòa tan x (g) KNO3 tạo thành dung dịch bão hòa.
\(x=\dfrac{106.5\cdot31.6}{100}=33.654\left(g\right)\)
\(n_{KNO_3}=\dfrac{33.654}{101}=0.33\left(mol\right)\)
\(m_{KNO_3\cdot2H_2O}=0.33\cdot\left(101+2\cdot18\right)=45.21\left(g\right)\)
PTHH :
\(Na_2O+H_2O\rightarrow2NaOH\)
0,04 0,08
\(n_{NaOH}=0,2.0,4=0,08\left(mol\right)\)
\(m_{Na_2O}=0,04.62=2,48\left(g\right)\)
Do CuO ko tác dụng với H2O nên 4gam chất rắn = KL CuO
Ta có :
m = 2,48 + 4 = 6,48 (g)
a, \(n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\)
PTHH: Na2O + H2O ---> 2NaOH
0,2------------------>0,4
\(\Rightarrow C\%_{NaOH}=\dfrac{0,4.40}{12,4+50}.100\%=25,64\%\)
b, \(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)
PTHH: 2Na + 2H2O ---> 2NaOH + H2
0,2------------------->0,2------->0,1
\(\Rightarrow C\%_{NaOH}=\dfrac{0,2.40+16}{100+16+4,6-0,1.2}.100\%==20\%\)
c, \(n_{Na}=\dfrac{9,2}{23}=0,4\left(mol\right)\)
\(n_{HCl}=\dfrac{100.7,3\%}{36,5}=0,2\left(mol\right)\)
PTHH:
2Na + 2HCl ---> 2NaCl + H2
0,2<-----0,2-----------0,2--->0,1
2Na + 2H2O ---> 2NaOH + H2
0,2------------------>0,2----->0,1
\(\Rightarrow m_{dd}=9,2+100-\left(0,1+0,1\right).2=108,8\left(g\right)\\ \Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{0,2.58,5}{108,8}.100\%=10,75\%\\C\%_{NaOH}=\dfrac{0,2.40}{108,8}.100\%=7,35\%\end{matrix}\right.\)
\(n_{Na_2O}=\dfrac{a}{62}\left(mol\right)\)
\(m_{NaOH}=\dfrac{400.1,15}{100}=4,6\left(g\right)\)
PTHH: Na2O + H2O --> 2NaOH
\(\dfrac{a}{62}\)------------->\(\dfrac{a}{31}\)
=> \(m_{NaOH}=\dfrac{40a}{31}=4,6\left(g\right)\) => a = 3,565 (g)
=> mdd sau pư = a + m = 400 (g) => m = 396,435 (g)