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![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1: \(Zn\left(0,04\right)+2HCl\left(0,08\right)\rightarrow ZnCl_2\left(0,04\right)+H_2\left(0,04\right)\)
\(n_{H_2}=0,04\left(mol\right)\)
\(TheoPTHH:n_{Zn}=0,04\left(mol\right)\Rightarrow m_1=2,6\left(g\right)\)
\(n_{HCl}=0,08\left(mol\right)\Rightarrow m_{HCl}=2,92\left(g\right)\)
\(\Rightarrow m_2=\dfrac{2,92.100}{14,6}=20\left(g\right)\)
\(mddsau=2,6+20-0,04.2=22,52(g)\)
\(TheoPTHH:n_{ZnCl_2}=0,04\left(mol\right)\)
\(\Rightarrow C\%_{ZnCl_2}=\dfrac{0,04.136.100}{22,52}=24,26\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
Sửa đề: 250ml NaCl 2 mol/l
Ta có: \(n_{NaCl}=0,25\cdot2=0,5\left(mol\right)\) \(\Rightarrow C_{M_{NaCl\left(sau\right)}}=\dfrac{0,5}{0,15+0,25}=1,25\left(M\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : nNa \(=\dfrac{4,6}{23}=0,2\left(mol\right)\)
PTHH : \(2Na+2H_2O\rightarrow2NaOH+H_2\uparrow\)
a) Theo pt :\(n_{H_2}=\dfrac{1}{2}.n_{Na}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
b) Theo pt : \(n_{NaOH}=n_{Na}=0,2\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,2.40=8\left(g\right)\)
\(m_{ddspu}=4,6+100-0,1.2=104,4\left(g\right)\)
\(\Rightarrow C\%_{NaOH}=\dfrac{8}{104,4}.100\%=7,66\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(n_{CuSO_4}=dfrac{8}{160}=0,05left(mol ight))
(V_{CuSO_4}=dfrac{m_{CuSO_4}}{D_{CuSO_4}}=dfrac{8}{1,26}=dfrac{400}{63}left(ml ight))
(m_{H_2O}=1.100=100left(g ight))
(n_{H_2O}=dfrac{100}{18}=dfrac{50}{9}left(mol ight))
(C_M=dfrac{0,05+dfrac{50}{9}}{dfrac{400}{63}+100}simeq0,05M)
(C_{\%}=dfrac{8}{100}.100\%=8\%)
Chết làm sai chỗ CM rồi
Để sửa lại nha, các bác đừng vội ném đá
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(C\%_{NaOH}=\dfrac{8}{8+100}.100\%\approx7,41\%\)
\(n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)
V dd sau hòa tan = 100 (ml) = 0,1 (l)
\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,2}{0,1}=2\left(M\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(C\%_{KCl}=\dfrac{20}{20+60}.100\%=25\%\)
b, \(C\%=\dfrac{40}{40+150}.100\%\approx21,05\%\)
c, \(C\%_{NaOH}=\dfrac{60}{60+240}.100\%=20\%\)
d, \(C\%_{NaNO_3}=\dfrac{30}{30+90}.100\%=25\%\)
e, \(m_{NaCl}=150.60\%=90\left(g\right)\)
f, \(m_{ddA}=\dfrac{25}{10\%}=250\left(g\right)\)
g, \(n_{NaOH}=120.20\%=24\left(g\right)\)
Gọi: nNaOH (thêm vào) = a (g)
\(\Rightarrow\dfrac{a+24}{a+120}.100\%=25\%\Rightarrow a=8\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\\ C_{M\left(HCl\right)}=\dfrac{0,2}{0,4}=0,5M\\ b,n_{H_2SO_4}=\dfrac{73,5}{98}=0,75\left(mol\right)\\ C_{M\left(H_2SO_4\right)}=\dfrac{0,75}{0,5}=1,5M\\ n_{NaOH}=\dfrac{16}{40}=0,4\left(mol\right)\\ C_{M\left(NaOH\right)}=\dfrac{0,4}{0,25}=1,6M\\ n_{Ba\left(OH\right)_2}=\dfrac{34,2}{171}=0,2\left(mol\right)\\ C_{M\left(Ba\left(OH\right)_2\right)}=\dfrac{0,2}{0,8}=0,25M\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Al}=\dfrac{2.7}{27}=0.1\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(0.1........0.3.........0.1...........0.15\)
\(m_{dd_{HCl}}=97.8\cdot1=97.8\left(g\right)\)
\(m_{ddsaupư}=2.7+97.8-0.15\cdot2=100.2\left(g\right)\)
\(C\%_{AlCl_3}=\dfrac{0.1\cdot133.5}{100.2}\cdot100\%=13.32\%\)
vậy phải kí hiệu là D mới đúng á bạn!
Bài làm
m(NaOh)= 8g => n(Naoh)= m/M= 8/40=0,2 mol
m(H2O)=32g; D(H2O)= 1g/ml; D( dd Naoh)=1,25g/ml
Giải
m(dd)= 8+32=40g
C%=8/32 x 100=25%
*( phần này là chắc chắn đúng)*
V(dd)=40l( vì 40g=40l)
Cm= n/V=0,2/40=0,005M
( phần này không chắc)
d này là khối lượng riêng hay là trọng lượng riêng vậy bạn?