nCuSO4.5H2O=\(\dfrac{50}{250}\)=0,2 mol

→nCuSO4=0,2(mol)

nH2O=0,2.5=1(mol)

mH2O=1.18=18(g)

VH2O=390+18=408(ml)

CMCuSO4=\(\dfrac{0,2}{0,408}=0,49M\)

mdd=50+390=440(g)

C%CuSO4=\(\dfrac{0,2.160}{440}100=7,27\%\)

tk

nCuSO4.5H2O==0,2(mol)

nCuSO4=0,2(mol)

nH2O=0,2.5=1(mol)

mH2O=1.18=18(g)

VH2O=390+18=408(ml)

CMCuSO4==0,49(M)

mdd=50+390=440(g)

C%CuSO4==7,27%

\(a,C\%_{CuSO_4}=\dfrac{5}{200+5}.100\%=2,43\%\\ b,C\%_{NaOH}=\dfrac{0,2.40}{200}.100\%=4\%\\ c,n_{NH_3}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ C\%_{NH_3}=\dfrac{0,3.17}{200+0,3.17}.100\%=2,5\%\\ d,n_{KCl}=\dfrac{9.10^{22}}{6.10^{23}}=0,15\left(mol\right)\\ C\%_{KCl}=\dfrac{0,15.74,5}{200}=5,5875\%\)

Ta có: \(n_{CuSO_4}=n_{CuSO_4.5H_2O}=\dfrac{25}{250}=0,1\left(mol\right)\)

m _{dd sau hòa tan} = 25 + 375 = 400 (g)

\(\Rightarrow C\%_{CuSO_4}=\dfrac{0,1.160}{400}.100\%=4\%\)

\(n_{CuSO_4}=n_{CuSO_4.5H_2O}=\dfrac{75}{250}=0,3\left(mol\right)\\ \rightarrow C_{M\left(CuSO_4\right)}=\dfrac{0,3}{0,9}=\dfrac{1}{3}M\)

Câu 1 : 

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

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

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

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

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

1

\(a)m_{H_2O}=250-5=245g\\b )C_{\%NaCl}=\dfrac{5}{250}\cdot100=2\%\)

\(2\\ m_{ddCuSO_4}=\dfrac{15.100}{5}=300g\\ m_{H_2O}=300-15=285g\)

Câu 1:

a, Ta có: m _dd = m _{chất tan} + m_H2O ⇒ m_H2O = 250 - 5 = 245 (g)

b, \(C\%_{NaCl}=\dfrac{5}{250}.100\%=2\%\)

Câu 2:

Ta có: \(C\%_{CuSO_4}=\dfrac{15}{m_{ddCuSO_4}}.100\%=5\%\)

\(\Rightarrow m_{ddCuSO_4}=300\left(g\right)\)

⇒ m_H2O = 300 - 15 = 285 (g)

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

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