Câu 4:

\(a,CaCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Ca+CO_2+H_2O\\ n_{CH_3COOH}=0,2.1=0,2\left(mol\right)\\ b,n_{CaCO_3}=n_{\left(CH_3COO\right)_2Ca}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ m_{CaCO_3}=0,1.100=10\left(g\right)\\ c,m_{muối}=m_{\left(CH_3COO\right)_2Ca}=158.0,1=15,8\left(g\right)\)

cứu mìn ae oii

TK nhé:

https://www.vietjack.com/vth-khoa-hoc-tu-nhien-6-kn/bai-16-7-trang-40-vth-khoa-hoc-tu-nhien-6-ket-noi.jsp#:~:text=Nh%C5%A9%20t%C6%B0%C6%A1ng%3A%20s%E1%BB%AFa%2C%20h%E1%BB%97n%20h%E1%BB%A3p,%2C%20n%C6%B0%E1%BB%9Bc%20b%C3%B9n%2C...

\(n_{CH_3COOH}=0,2.1=0,2\left(mol\right);n_{Ca\left(OH\right)_2}=0,2.1=0,2\left(mol\right)\\ a,PTHH:Ca\left(OH\right)_2+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Ca+2H_2O\\ Vì:\dfrac{0,2}{1}>\dfrac{0,2}{2}\Rightarrow Ca\left(OH\right)_2dư\\ n_{\left(CH_3COO\right)_2Ca}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ m_{muối}=m_{\left(CH_3COO\right)_2Ca}=158.0,1=15,8\left(g\right)\\ b,Ca\left(OH\right)_2dư\Rightarrow Quỳ.tím.chuyển.màu.xanh\)

c, Vì dư Ca(OH)₂ nên cần thêm axit vào để trung hoà lượng bazo dư, ta dùng dd CH₃COOH

\(n_{CH_3COOH\left(thêm\right)}=2.n_{Ca\left(OH\right)_2\left(dư\right)}=2.\left(0,2-\dfrac{0,2}{2}\right)=0,2\left(mol\right)\\ V_{ddCH_3COOH\left(thêm\right)}=\dfrac{0,2}{1}=0,2\left(l\right)=200\left(ml\right)\)

a, $NaCl+AgN{O}_3\to NaN{O}_3+AgC{l}_{\downarrow }$

b, Không pư.

c, $Fe+CuC{l}_2\to FeC{l}_2+Cu$

d, $BaC{l}_2+H_{2}S{O}_4\to 2HCl+BaS{O}_{4\downarrow }$

e, Không pư.

a: \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)

b: \(n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)

=>\(n_{CH_3COO_H}=0,2\left(mol\right)\)

500ml=0,5(lít)

\(C_{M\left(CH_3COOH\right)}=\dfrac{0.2}{0.5}=0,4\)