\(n_{Ca\left(OH\right)_2}=0,25.0,2=0,05\left(mol\right)\)

\(n_{Ca}=\frac{8}{40}=0,2\left(mol\right)\)

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

PTHH: \(Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\)

______0,2-------------------->0,2___________(mol)

\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

0,1-------------------->0,1___________(mol)

=> \(n_{Ca\left(OH\right)_2}=0,1+0,2+0,05=0,35\left(mol\right)\)

=> \(C_M=\frac{0,35}{0,25}=1,4M\)

\(a.n_{CuCl_2}=\dfrac{13,5.10\%}{100\%.135}=0,01mol\\ CuCl_2+Ca\left(OH\right)_2\rightarrow Cu\left(OH\right)_2+CaCl_2\\ n_{Ca\left(OH\right)_2}=n_{Cu\left(OH\right)_2}=n_{CaCl_2}=0,01mol\\ m_{ddCa\left(OH\right)_2}=\dfrac{0,01.74}{25\%}\cdot100\%=2,96g\\ b.m_{Cu\left(OH\right)_2}=0,01.98=0,98g\\ c.m_{dd}=13,5+2,96-0,98=15,48g\\ C_{\%CaCl_2}=\dfrac{0,01.111}{15,48}\cdot100\%=7,17\%\\ d.Cu\left(OH\right)_2\xrightarrow[]{t^0}CuO+H_2O\\ n_{CuO}=n_{Cu\left(OH\right)_2}=0,01mol\\ m_{CuO}=0,01.80=0,8g\)

\(Đặt:n_{Na_2CO_3}=a\left(mol\right);n_{K_2CO_3}=b\left(mol\right)\\ Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\\ K_2CO_3+2HCl\rightarrow2KCl+CO_2+H_2O\\ CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+H_2O\\ n_{CaCO_3}=n_{CO_2}=0,3\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}106a+138b=38,2\\a+b=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\\ a.C\%_{ddHCl}=\dfrac{0,6.36,5}{200}.100=10,95\%\\ b.m_{ddB}=38,2+200-0,3.44=225\left(g\right)\\ C\%_{ddKCl}=\dfrac{74,5.2.0,2}{225}.100\approx13,244\%\\ C\%_{ddNaCl}=\dfrac{58,5.2.0,1}{225}.100=5,2\%\)

trả lời giùm mik vs

SO2 + Ca(OH)2 -> CaSO3 + H2O

a................................a (mol)

với a = V/22,4 (mol)

Theo bài ra :

\(\Delta m=m_{CáSO3}-m_{SO2}=120a-64a=56a\) = 5,6 (g)

=> a = 0,1(mol)

=> V =2,24(l)

Gọi số mol SO2 là x(mol)

SO2+ Ca(OH)2---> CaSO3+ H2O

x...................................x.(mol)

Có: mCaSO3-mSO2=56x=5,6(g)

=> x=0,1(mol)

=> V=0,1.22,4=2,24(l)

\(\text{Quy đổi hỗn hợp gồm : Na , Ca , O }\)

\(n_{Na}=n_{NaOH}=0.4\left(mol\right)\)

\(n_{Ca\left(OH\right)_2}=0.25\left(mol\right)\)

\(m_O=23.2-0.4\cdot23-0.25\cdot40=4\left(g\right)\)

\(n_O=\dfrac{4}{16}=0.25\left(mol\right)\)

\(Na\rightarrow Na^++1e\)

\(Ca\rightarrow Ca^{+2}+2e\)

\(O+2e\rightarrow O^{2-}\)

\(2H^{+1}+2e\rightarrow H_2^0\)

\(\text{Bảo toàn electron : }\)

\(n_{Na}+2n_{Ca}=2n_O+2n_{H_2}\)

\(\Rightarrow0.4+2\cdot0.25=2\cdot0.25+2\cdot n_{H_2}\)

\(\Rightarrow n_{H_2}=0.2\left(mol\right)\)

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

Đáp án D

(a) Đun sôi nước cứng tạm thời.

(b) Cho phèn chua vào lượng dư dd Ba(OH)₂.

(c) Cho dd NH₃ dư vào dd AlCl_3.