C_M (mol/l) chứ nhỉ đề cho mol rồi mà

\(a,C_{M\left(NaOH\right)}=\dfrac{n}{V}=\dfrac{0,2}{0,2}=1M\\ b,n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\\ C_{M\left(HCl\right)}=\dfrac{n}{V}=\dfrac{0,2}{0,5}=0,4M\\ c,n_{NH_3}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ C_{M\left(NH_3\right)}=\dfrac{0,3}{0,3}=1M\\ d,n_{H_2SO_4}=\dfrac{4,9}{98}=0,05\left(mol\right)\\ C_{M\left(H_2SO_4\right)}=\dfrac{0,05}{0,25}=0,2M\)

a)\(C\%=\dfrac{m_{ct}}{m_{dd}}\cdot100\%=\dfrac{5}{5+45}\cdot100\%=10\%\)

b)\(C\%=\dfrac{m_{ct}}{m_{dd}}\cdot100\%=\dfrac{5,6}{5,6+94,4}\cdot100\%=5,6\%\)

c)\(m_{ctNaOH}=\dfrac{200\cdot10\%}{100\%}=20g\)

   \(m_{ctNaOH}=\dfrac{300\cdot5\%}{100\%}=15g\)

\(C\%=\dfrac{m_{ct}}{m_{dd}}\cdot100\%=\dfrac{20+15}{200+300}\cdot100\%=7\%\)

\(a,C\%_{NaOH}=\dfrac{5}{5+45}=10\%\)

b, \(n_{CaO}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PTHH: CaO + H₂O ---> Ca(OH)₂

              0,1 ---------------> 0,1

\(\rightarrow C\%_{Ca\left(OH\right)_2}=\dfrac{74.0,1}{5,6+94,4}=37\%\)

c, \(m_{NaOH}=10\%.200+5\%.300=35\left(g\right)\)

\(\rightarrow C\%_{NaOH}=\dfrac{35}{200+300}=7\%\)

Câu 1 : 

\(n_{Al}=\dfrac{2.7}{27}=0.1\left(mol\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(0.1......................0.1.........0.15\)

\(m_{dd}=2.7+200-0.15\cdot2=202.4\left(g\right)\)

\(C\%_{AlCl_3}=\dfrac{0.1\cdot133.5}{202.4}\cdot100\%=6.59\%\)

\(a,n_{CaCO_3}=\dfrac{30}{100}=0,3mol\\ b,S_{NaCl}=\dfrac{72}{200}\cdot100=36g\)

a, \(n_{CaCO_3}=\dfrac{30}{100}=0,3\left(mol\right)\)

b, \(S=\dfrac{72}{200}.100=36\left(g\right)\)

\(n_{CO_2}=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ n_{NaOH}=\dfrac{m}{M}=\dfrac{m_{dd}.C\%}{M}=\dfrac{200.16\%}{40}=0,8\left(mol\right)\)

Có: \(\dfrac{n_{NaOH}}{n_{CO_2}}=4\)

=> Phản ứng tạo muối Na₂CO₃

PT:

\(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

0,2.       0,4            0,2

=> dd sau phản ứng có những chất tan là:

\(\left\{{}\begin{matrix}Na_2CO_3:0,2\left(mol\right)\\NaOH:0,4\left(mol\right)\end{matrix}\right.\)

m_dd spu=0,2.44+200=208,8(g)

\(\%m_{NaOH}=\dfrac{0,4.40}{208,8}.100\%=7,66\%\\\%m_{Na_2CO_3}=\dfrac{0,2.106}{208,8}.100\%=10,15\% \)

Câu 1:

nAl= 0,1(mol)

PTHH: 2Al + 6 HCl -> 2 AlCl3 + 3 H2 

nAlCl3= nAl=0,1(mol)

-> mAlCl3= 133,5 x 0,1= 13,35(g)

mddAlCl3= mAl + mddHCl - mH2 = 2,7 + 200 - 3/2 x 0,1 x 2= 202,4(g)

C%ddAlCl3= (13,35/202,4).100= 6,596%

bạn ơi, có nồng độ của dd HCl không? vì nếu làm ntn thì câu 1 đề bài không chặt chẽ

\(a,m_{dd}=\dfrac{5}{12\%}=\dfrac{125}{3}\left(g\right)\\ b,m_{dd}=\dfrac{4}{7,3\%}=\dfrac{4000}{73}\left(g\right)\\ c,m_{NaOH}=0,5.40=20\left(g\right)\\ m_{dd}=\dfrac{20}{10\%}=200\left(g\right)\)

\(m_{\text{dd}}=\dfrac{5.100}{12}=41,6\left(g\right)\\ m_{\text{dd}}=\dfrac{4.100}{7,3}=\dfrac{4000}{73}\left(g\right)\\ m_{\text{dd}}=\dfrac{\left(0,5.40\right).100}{10}=200\left(g\right)\)

a) \(m_{C_{12}H_{22}O_{11}\left(cần.lấy\right)}=\dfrac{10.200}{100}=20\left(g\right)\)

=> dd chưa bão hòa

b) \(m_{NaCl\left(cần.lấy\right)}=\dfrac{10.36}{100}=3,6\left(g\right)\)

=> dd đã bão hòa 

c) Gọi khối lượng KBr là a (g)

=> m_H2O = 330 - a (g)

Có: \(S=\dfrac{a}{330-a}.100=120\left(g\right)\)

=> a = 180 (g)

=> m_H2O = 330 - 180 = 150 (g)

d) \(m_{NaCl}=\dfrac{500.36}{100}=180\left(g\right)\)

a, Xét \(\dfrac{18,8}{10}.100=188\rightarrow\) dd chưa bão hoà

b, Xét \(\dfrac{4,2}{10}.100=42\left(g\right)\rightarrow\) đã bão hoà và còn dư muối

c, Gọi \(m_{KBr}=a\left(g\right)\)

\(\rightarrow S_{KBr\left(60^oC\right)}=\dfrac{a}{330-a}.100=120\\ \rightarrow\left\{{}\begin{matrix}m_{KBr}=180\left(g\right)\\m_{H_2O}=330-180=150\left(g\right)\end{matrix}\right.\)

d, Gọi \(m_{NaCl}=a\left(g\right)\)

\(\rightarrow S_{NaCl\left(25^oC\right)}=\dfrac{a}{500-a}.100=36\\ \rightarrow m_{NaCl}=132,35\left(g\right)\)

eyyyyyy, cảm ơn bro nhé. 

\(m_{ddBaCl_2}=15+200=215\left(g\right)\)

\(C\%=\dfrac{15}{215}.100\%\approx6,98\%\)