a) \(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)

\(n_{HCl}=\dfrac{146.5\%}{36,5}=0,2\left(mol\right)\)

PTHH: CuO + 2HCl --> CuCl₂ + H₂O

Xét tỉ lệ: \(\dfrac{0,05}{1}< \dfrac{0,2}{2}\) => CuO hết, HCl dư

=> dd sau phản ứng chứa CuCl₂, HCl dư

b) 

PTHH: CuO + 2HCl --> CuCl₂ + H₂O

           0,05-->0,1------>0,05

m_{dd sau pư} = 4 + 146 = 150 (g)

\(\left\{{}\begin{matrix}C\%_{CuCl_2}=\dfrac{0,05.135}{150}.100\%=4,5\%\\C\%_{HCldư}=\dfrac{\left(0,2-0,1\right).36,5}{150}.100\%=2,433\%\end{matrix}\right.\)

b) 

PTHH: NaOH + HCl --> NaCl + H₂O

            CuCl₂ + 2NaOH --> 2NaCl + Cu(OH)₂

             0,05--------------------------->0,05

             Cu(OH)₂ --t^o--> CuO + H₂O

               0,05----------->0,05

=> \(a=m_{Cu\left(OH\right)_2}=0,05.98=4,9\left(g\right)\)

=> \(b=m_{CuO}=0,05.80=4\left(g\right)\)

\(m_{HCl}=100.7,3\%=7,3\left(g\right)\Rightarrow n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)

PT: \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

Theo PT: \(n_{BaCl_2}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\)

\(\Rightarrow C\%_{BaCl_2}=\dfrac{0,1.208}{100+100}.100\%=10,4\%\)

HCl k có phần trăm à

Ta có: \(n_{ZnO}=\dfrac{16,2}{81}=0,2\left(mol\right)\)

a, PT: \(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

______0,2_____0,4____0,2 (mol)

b, \(m_{HCl}=0,4.36,5=14,6\left(g\right)\Rightarrow m_{ddHCl}=\dfrac{14,6}{10\%}=146\left(g\right)\)

c, \(C\%_{ZnCl_2}=\dfrac{0,2.136}{16,2+146}.100\%\approx16,77\%\)

Bạn tham khảo nhé!

\(n_{CuSO_4}=\dfrac{160.10\%}{160}=0,1\left(mol\right)\)

\(n_{NaOH}=\dfrac{150.8\%}{40}=0,3\left(mol\right)\)

PTHH: CuSO₄ + 2NaOH --> Cu(OH)₂ + Na₂SO₄

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,3}{2}\) => CuSO₄ hết, NaOH dư

PTHH: CuSO₄ + 2NaOH --> Cu(OH)₂ + Na₂SO₄

                0,1------>0,2------->0,1------->0,1

=> m = 0,1.98 = 9,8 (g)

\(\left\{{}\begin{matrix}m_{NaOH_{dư}}=\left(0,3-0,2\right).40=4\left(g\right)\\m_{Na_2SO_4}=0,1.142=14,2\left(g\right)\end{matrix}\right.\)

m_{dd sau pư} = 160 + 150 - 9,8 = 300,2 (g)

\(\left\{{}\begin{matrix}C\%_{NaOH_{dư}}=\dfrac{4}{300,2}.100\%=1,33\%\\C\%_{Na_2SO_4}=\dfrac{14,2}{300,2}.100\%=4,73\%\end{matrix}\right.\)

\(m_{CuSO_4}=\dfrac{160.10}{100}=16\left(g\right)\\ n_{NaOH}=\dfrac{8.150}{100}=12\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}n_{CuSO_4}=\dfrac{16}{160}=0,1\left(mol\right)\\n_{NaOH}=\dfrac{12}{40}=0,3\left(mol\right)\end{matrix}\right.\)

PTHH: 2NaOH + CuSO₄ ---> Cu(OH)₂ + Na₂SO₄

LTL: \(0,1< \dfrac{0,3}{2}\rightarrow\) NaOH dư

Theo pt: \(\left\{{}\begin{matrix}n_{NaOH\left(pư\right)}=\dfrac{1}{2}n_{CuSO_4}=2.0,1=0,2\left(mol\right)\\n_{Na_2SO_4}=n_{Cu\left(OH\right)_2}=n_{CuSO_4}=0,1\left(mol\right)\end{matrix}\right.\)

\(\rightarrow m=0,1.98=9,8\left(g\right)\\ m_{dd}=160+150-9,8=300,2\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}C\%_{NaOH\left(dư\right)}=\dfrac{\left(0,3-0,2\right).40}{300,2}=1,33\%\\C\%_{Na_2SO_4}=\dfrac{0,1.142}{300,2}=4,73\%\end{matrix}\right.\)

\(CaCO_3+ HCl → CaCl_2+H_2O +CO_2\)

\(n_{CaCO_3}=\dfrac{10}{40+12+16.3}=0,1(mol)\)

\(n_{HCl}=\dfrac{146}{1+35,5}=4(mol)\)

\(\Rightarrow n_{HCl_{dư}}=4-0,1=3,9(mol) ; n_{CaCl_2}=0,1(mol)\\\Rightarrow m_{\text{chất tan}} = m_{HCl_{dư}}+m_{CaCl_2}\\=0,39.(35,5+1)+0,1(40+35,5.2)=25,335(g)\)

Vậy...

PT: \(CaCO_3+2HCl→CaCl_2+H_2O+CO_2\)

\(n_{HCl_{dư}} = 4-0,1.2=3,8(mol) \Rightarrow m_{HCl_{dư}}=138,7 (g) \\ n_{CaCl_2}=0,1(mol) \Rightarrow m_{CaCl_2}=11,1(g) \\ m_{\text{chất tan}}=149,8 (g) \\ \Rightarrow C\%=\dfrac{149,8}{146+10} .100\%=96\%\).

a)m_H2SO4=\(\dfrac{200.7,3\text{%}}{100\%}\)=14,6g

n_HCl=\(\dfrac{14,6}{36,5}\)=0,4(mol)

PTHH:

NaOH+ HCl→ NaCl+ H₂O

1            1            1         1

0,4        0,4         0,4     (mol)

⇒m_NaOH=0,4.40=16(g)

Nồng độ % của dd NaOH cần dùng là:

C_%NaOH=\(\dfrac{16}{200}\)   .100%=8%

b)Ta có:m_{dd spứ}=m_{dd trc pứ}=400g

m_NaCl=0,4.58,5=23,4g

Nồng độ % dd muối tạo thành sau pứ là:

C_%dd NaCl=\(\dfrac{23,4}{400}\)   .100%=5,85%        

a) \(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)

PTHH: CuO + 2HCl --> CuCl₂ + H₂O

           0,05---->0,1----->0,05--->0,05

=> \(C\%\left(HCl\right)=\dfrac{0,1.36,5}{150}.100\%=2,433\%\)

b) \(C\%\left(CuCl_2\right)=\dfrac{0,05.135}{4+150}.100\%=4,383\%\)

Giải:

Số mol CaCO3 và HCl lần lượt là:

nCaCO3 = m/M = 10/100 = 0,1 (mol)

nHCl = (mdd.C%)/(100.M) = (200.5)/(100.36,5) = 0,3 (mol)

Tỉ lệ: nCaCO3 : nHCl = 0,1/1 : 0,3/2 = 0,1 : 0,15

=> HCl dư, tính theo CaCO3

=> mHCl(dư) = n(dư).M = 0,1.36,5 = 3,65 (g)

PTHH: CaCO3 + 2HCl -> CaCl2 + H2CO3

----------0,1----------0,3--------0,1----0,1----

Khối lượng dd sau phản ứng là:

mddspư = mCaCO3 + mddHCl = 10 + 200 = 210 (g)

Nồng độ phần trăm các chất trong dd sau phản ứng là:

C%HCl(dư) = (mct/mddspư).100 = (3,65/210).100 ≃ 1,74 %

C%CaCl2 = (mct/mddspư).100 = (111.0,1/210).100 ≃ 5,3 %

C%H2CO3 = (mct/mddspư).100 = (62.0,1/210).100 ≃ 3 %

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