\(M_2CO_3+2HCl\rightarrow2MCl+H_2O+CO_2\)

a---------->2a--------------------------->a

\(MHCO_3+HCl\rightarrow MCl+H_2O+CO_2\)

b---------->b---------------------------->b

\(NaOH+HCl\rightarrow NaCl+H_2O\)

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

\(\left\{{}\begin{matrix}a+b=n_{CO_2}=0,3\\2a+b+0,1=0,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\)

=> \(0,1\left(2M+60\right)+0,2\left(M+61\right)=27,4\Rightarrow M=23\)

M là Na

Hai muối ban đầu là \(Na_2CO_3,NaHCO_3\)

\(m_{Na_2CO_3}=0,1.106=10,6\left(g\right)\)

\(m_{NaHCO_3}=0,2.84=16,8\left(g\right)\)

b. Trong đề không có đề cập tới V bạn.

R+HCln->RCln+H:

\(2R+2nHCl\rightarrow2RCl_n+nH_2\)

1. Gọi CTHH của oxit là N_xO_y.

Ta có: \(\dfrac{m_N}{m_O}=\dfrac{7}{20}\Rightarrow\dfrac{n_N}{n_O}=\dfrac{7}{20}:\dfrac{14}{16}=\dfrac{2}{5}\)

⇒ x:y = 2:5

→ N₂O₅

2. Gọi CTHH cần tìm là Fe_xO_y.

\(\Rightarrow\dfrac{m_{Fe}}{m_O}=\dfrac{7}{2}\Rightarrow\dfrac{n_{Fe}}{n_O}=\dfrac{7}{2}:\dfrac{56}{16}=1\)

⇒ x:y = 1

→ FeO

3. CTHH cần tìm: RO₂

Mà: %R = 46,7%
\(\Rightarrow\dfrac{M_R}{M_R+16.2}.100\%=46,7\%\)

⇒ M_R = 28 (g/mol)

→ SiO₂

Ta có: 40n_NaOH + 56n_KOH = 3,04 (1)

PT: \(NaOH+HCl\rightarrow NaCl+H_2O\)

\(KOH+HCl\rightarrow KCl+H_2O\)

Theo PT: \(n_{NaCl}=n_{NaOH}\)

\(n_{KCl}=n_{KOH}\)

⇒ 58,5n_NaOH + 74,5n_KOH = 4,15 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{NaOH}=0,02\left(mol\right)\\n_{KOH}=0,04\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\%m_{KOH}=\dfrac{0,04.56}{3,04}.100\%\approx73,68\%\)

→ Đáp án: C

a

PT: \(Cu+2AgNO_3\rightarrow Cu\left(NO_3\right)_2+2Ag\)

a, m _{AgNO3 (pư)} = 250.17%.6% = 2,55 (g)

\(\Rightarrow n_{AgNO_3\left(pư\right)}=\dfrac{2,55}{170}=0,015\left(mol\right)\)

Theo PT: n_{Cu (pư)} = 1/2n_AgNO3 = 0,0075 (mol)

n_Ag = n_AgNO3 = 0,015 (mol)

⇒ m _{vật lấy ra} = 50 - m_{Cu (pư)} - m_Ag = 51,14 (g)

b, Ta có: m _{dd sau pư} = 0,0075.64 + 250 - 0,015.108 = 248,86 (g)

Theo PT: n_Cu(NO3)2 = 1/2n_AgNO3 = 0,0075 (mol)

\(\Rightarrow C\%_{Cu\left(NO_3\right)_2}=\dfrac{0,0075.188}{248,86}.100\%\approx0,57\%\)

\(C\%_{AgNO_3}=\dfrac{250.6\%-2,55}{248,86}.100\%\approx5\%\)

\(a,CaO+H_2O\rightarrow Ca\left(OH\right)_2\\ m_{rắn}=m_{MgO}=4g\\ m_{CaO}=9,6-4=5,6g\\ c,n._{CaO}=\dfrac{5,6}{56}=0,1mol\\ C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,1}{0,2}=0,5M\)

Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

a, n_H2 = n_Zn = 0,2 (mol)

⇒ V_H2 = 0,2.24,79 = 4,958 (l)

b, n_ZnCl2 = n_Zn = 0,2 (mol)

⇒ m_ZnCl2 = 0,2.136 = 27,2 (g)

c, \(H=\dfrac{3,225}{4,958}.100\%\approx65,05\%\)

Nồng độ %? hoặc 500 ml dd HCl

\(n_{CaCO_3}=\dfrac{30}{100}=0,3mol\\ CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ n_{HCl}=0,3.2=0,6mol\\ \left[{}\begin{matrix}C_{M_{HCl}}\\C_{\%HCl}=\dfrac{0,6.36,5}{500}\cdot100=4,38\%\end{matrix}\right.=\dfrac{0,6}{0,5}=1,2M}\)