Ta có: \(n_{Cu\left(NO_3\right)_2}=0,2.1,5=0,3\left(mol\right)\)

PT: \(Cu\left(NO_3\right)_2+2NaOH\rightarrow Cu\left(OH\right)_{2\downarrow}+2NaNO_3\)

_______0,3_______0,6_______0,3_________0,6 (mol)

a, m_Cu(OH)2 = 0,3.98 = 29,4 (g)

b, \(V_{ddNaOH}=\dfrac{0,6}{2}=0,3\left(l\right)\)

c, \(C_{M_{NaNO_3}}=\dfrac{0,6}{0,2+0,3}=1,2M\)

Bạn tham khảo nhé!

a) \(n_{Cu\left(NO_3\right)_2}=1,5.0,2=0,3\left(mol\right)\)

\(Cu\left(NO_3\right)_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl\)

\(n_{Cu\left(OH\right)_2}=n_{Cu\left(NO_3\right)_2}=0,3\left(mol\right)\)

=> \(m_{Cu\left(OH\right)_2}=29,4\left(g\right)\)

b) \(n_{NaOH}=2n_{Cu\left(OH\right)_2}=0,6\left(mol\right)\)

=> \(V_{NaOH}=\dfrac{0,6}{2}=0,3\left(l\right)\)

c) \(CM_{NaCl}=\dfrac{0,3.2}{0,2+0,3}=1,2M\)

\(3NaOH+FeCl_3\rightarrow Fe\left(OH\right)_3+3NaCl\)

\(n_{NaCl}=n_{NaOH}=0,2.3=0,6\left(mol\right)\)

=> \(C_{M\left(NaCl\right)}=\dfrac{0,6}{0,2}=3M\)

\(n_{Fe\left(ỌH\right)_3}=\dfrac{1}{3}n_{NaOH}=0,2\left(mol\right)\)

\(2Fe\left(OH\right)_3-^{t^o}\rightarrow Fe_2O_3+3H_2O\)

Ta có \(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_3}=0,1\left(mol\right)\)

=> m _Fe2O3 = 0,1 . 160=16(g)

a)

$K_2SO_4 + BaCl_2 \to BaSO_4 + 2KCl$

b)

$n_{K_2SO_4} = 0,2.2 = 0,4(mol)$
$n_{BaCl_2} = 0,3.1 = 0,3(mol)$

Ta thấy : 

$n_{K_2SO_4} : 1 > n_{BaCl_2} : 1$ nên $K_2SO_4$ dư

$n_{BaSO_4} = n_{BaCl_2} = 0,3(mol)$
$m_{BaSO_4} = 0,3.233 = 69,9(gam)$

c) $n_{K_2SO_4} = 0,4 - 0,3 = 0,1(mol)$

$V_{dd\ sau\ pư} = 0,2 + 0,3 = 0,5(lít)$

$C_{M_{K_2SO_4} } = \dfrac{0,1}{0,5} = 0,2M$
$C_{M_{KCl}} = \dfrac{0,6}{0,5} = 1,2M$

\(n_{H_2SO_4}=0,3.1=0,3\left(mol\right)\)

PTHH: 2NaOH + H₂SO₄ --> Na₂SO₄ + 2H₂O

_______0,6<------0,3----------->0,3

=> V = \(\dfrac{0,6}{1}=0,6\left(l\right)\)

b) \(C_{M\left(Na_2SO_4\right)}=\dfrac{0,3}{0,6+0,3}=0,333M\)

\(n_{H_2SO_4}=1.0,3=0,3(mol)\\ 2NaOH+H_2SO_4\to Na_2SO_4+2H_2O\\ \Rightarrow n_{NaOH}=0,6(mol)\\ a,V_{dd_{NaOH}}=\dfrac{0,6}{1}=0,6(l)\\ b,n_{Na_2SO_4}=0,3(mol)\\ \Rightarrow C_{M_{Na_2SO_4}}=\dfrac{0,3}{0,6+0,3}=0,33M\)

a, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(MgCl_2+2NaOH\rightarrow2NaCl+Mg\left(OH\right)_{2\downarrow}\)

\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)

b, Ta có: \(n_{Mg}=\dfrac{9,6}{24}=0,4\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Mg}=0,8\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,8}{0,2}=4\left(M\right)\)

c, Theo PT: \(n_{MgO}=n_{Mg}=0,4\left(mol\right)\)

\(\Rightarrow m_{MgO}=0,4.40=16\left(g\right)\)

\(n_{MgCl_2}=0,15.0,2=0,03(mol)\\ PTHH:MgCl_2+2NaOH\to Mg(OH)_2\downarrow +2NaCl\\ a,n_{Mg(OH)_2}=n_{MgCl_2}=0,03(mol)\\ \Rightarrow m_{\downarrow}=m_{Mg(OH)_2}=0,03.58=1,74(g)\\ b,n_{NaOH}=2n_{MgCl_2}=0,06(mol)\\ \Rightarrow C_{M_{NaOH}}=\dfrac{0,06}{0,3}=0,2M\\ c,PTHH:Mg(OH)_2\xrightarrow{t^o}MgO+H_2O\\ \Rightarrow n_{MgO}=n_{Mg(OH)_2}=0,03(mol)\\ \Rightarrow m_{A}=m_{MgO}=0,03.40=1,2(g)\)

Cảm ơn.

\(a,PTHH:CuCl_2+2KOH\rightarrow Cu\left(OH\right)_2\downarrow+2KCl\\ n_{CuCl_2}=\dfrac{27}{135}=0,2\left(mol\right)\\ \Rightarrow n_{KOH}=2n_{CuCl_2}=0,4\left(mol\right)\\ \Rightarrow C\%_{KOH}=\dfrac{0,4}{0,2}=2M\\ b,PTHH:Cu\left(OH\right)_2\rightarrow^{t^o}CuO+H_2O\\ \Rightarrow n_{CuO}=n_{Cu\left(OH\right)_2}=n_{CuCl_2}=0,2\left(mol\right)\\ \Rightarrow m_{CuO}=0,2\cdot80=16\left(g\right)\)

\(200ml=0,2l\\ n_{Na_2CO_3}=0,5.0,2=0,1\left(mol\right)\\ PTHH:Na_2CO_3+2HCl\rightarrow2NaCl+CO_2\uparrow+H_2O\\ \left(mol\right)........0,1\rightarrow...0,2.......0,2..........0,1.........0,1\\ a,C_{M_{HCl}}=\dfrac{0,2}{0,2}=1\left(M\right)\\ b,m_{NaCl}=0,2.58,5=11,7\left(g\right)\\c, V_{ddNaCl}=V_{ddNa_2CO_3}+V_{ddHCl}=0,2+0,2=0,4\left(l\right)\\ C_{M_{NaCl}}=\dfrac{0,2}{0,4}=0,5\left(M\right)\)

vì em trộn 2 dung dịch lại với nhau mà, ví dụ em đổ 1 chai nước 500ml vào 1 chai nước 500 ml thì mình phải được 1 lít nước chứ