\(a.n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ n_{HCl}=0,2.1,5=0,3\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ Vì:\dfrac{0,3}{2}< \dfrac{0,2}{1}\\ \Rightarrow Mgdư\\ n_{H_2}=n_{MgCl_2}=\dfrac{0,3}{2}=0,15\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\\ b.V_{ddsau}=V_{ddHCl}=0,2\left(l\right)\\ C_{MddMgCl_2}=\dfrac{0,15}{0,2}=0,75\left(M\right)\)

a) \(n_{\left(CH_3COO\right)_2Mg}=\dfrac{7,1}{142}=0,05\left(mol\right)\)

PTHH: 2CH₃COOH + Mg --> (CH₃COO)₂Mg + H₂

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

=> V_H2 = 0,05.22,4 = 1,12 (l)

b) \(C_{M\left(dd.CH_3COOH\right)}=\dfrac{0,1}{0,025}=4M\)

PTHH: \(CaCl_2+Na_2CO_3\rightarrow CaCO_3\downarrow+2NaCl\)

a+b) Ta có: \(n_{CaCl_2}=0,1\cdot2=0,2\left(mol\right)\) 

\(\Rightarrow\left\{{}\begin{matrix}n_{CaCO_3}=0,2\left(mol\right)\\n_{NaCl}=0,4\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaCO_3}=0,2\cdot100=20\left(g\right)\\C_{M_{NaCl}}=\dfrac{0,4}{0,1+0,2}\approx1,33\left(M\right)\end{matrix}\right.\)

c) PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\)

Theo PTHH: \(n_{HCl}=2n_{CaCO_3}=0,4\left(mol\right)\) \(\Rightarrow m_{ddHCl}=\dfrac{0,4\cdot36,5}{10\%}=146\left(g\right)\)

a.PTHH:CaCl2+Na2CO3--->CaCO3+2NaCl

Ta có:nCaCl2=0,2

=>nCaCO3=nCaCl2=0,2(mol)=>mCaCO3(kết tủa)=100.0,2=2(g)

b.Vdd=100+200=300(ml)=0,3(l)

CM Nacl=(2.0,2)/0,3=4/3(M)(Đề cho 2 chất td vừa đủ nên dd sau pứ chỉ có NaCl)

c.CaCO3+2HCl--->CaCl2+CO2+H2O

nHCl(cần dùng)=2.0.2=0,4(mol)=>mHCl=36,5.0,4=14,6(g)

=>mddHCl=14,6/10%=146(g)

a, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

PTHH: Zn + 2HCl ---> ZnCl₂ + H₂

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

V_H2 = 0,1.22,4 = 2,24 (l)

b, \(C_{M\left(HCl\right)}=\dfrac{0,2}{0,2}=1M\)

c, \(C_{M\left(ZnCl_2\right)}=\dfrac{0,1}{0,2}=0,5M\)

\(n_{Cl_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

\(2NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O\)

 2 mol        1 mol    1mol      1mol      1 mol

0,1          0,05         0,05         0,05      0,05

\(V_{NaOH}=\dfrac{0,1}{1}=0,1\left(l\right)\)

\(CM_{NaCl}=\dfrac{0,05}{0,1}=0,5M\)

\(CM_{NaClO}=\dfrac{0,05}{0,1}=0,5M\)

a)

$CuO + H_2SO_4 \to CuSO_4 + H_2O$

$n_{CuO} = 0,25(mol) < n_{H_2SO_4} = 0,4(mol)$ nên $H_2SO_4$ dư

$n_{CuSO_4} = n_{CuO} = 0,25(mol)$
$m_{CuSO_4} = 0,25.160 = 40(gam)$

b)

$n_{H_2SO_4\ dư}  = 0,4 - 0,25 = 0,15(mol)$
$C_{M_{CuSO_4}} = \dfrac{0,25}{0,2} = 1,25M$
$C_{M_{H_2SO_4\ dư}} = \dfrac{0,15}{0,2} = 0,75M$

a) nCuO= 0,25(mol); nH2SO4= 0,4(mol)

PTHH: CuO + H2SO4 -> CuSO4 + H2O

0,25/1 < 0,4/1

=> CuO hết, H2SO4 dư, tính theo nCuO.

=> nCuSO4=nCuO=nH2SO4(p.ứ)=0,25(mol)

=> mCuSO4=0,25.160=40(g)

b) nH2SO4(dư)=0,4-0,25=0,15(mol)

Vddsau=VddH2SO4=200(ml)=0,2(l)

=>CMddCuSO4=0,25/0,2=1,25(M)

CMddH2SO4(dư)=0,15/0,2=0,75(M)

a).CuO + H₂SO₄ -> CuSO₄ + H₂O.

0,25                       0,25

n_CuO= \(\dfrac{20}{80}\)=0,25 (mol)

=> n_CuSO4=n_CuO= 0,25 (mol)

=> m_CuSO4= n.M= 0,25.160=40(g)

b).                         200ml=0,2l

C_MddCuSO4= \(\dfrac{n}{V}\)=\(\dfrac{0,25}{0,2}\)= 1,25 (M)

Câu 1
\(a)PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ b)200ml=0,2l\\ n_{HCl}=0,2.1=0,2mol\\ n_{H_2}=n_{MgCl_2}=\dfrac{1}{2}n_{H_2}=\dfrac{1}{2}\cdot0,2=0,1mol\\ V_{H_2}=0,1.24,79=2,479l\\ c)C_{M_{MgCl_2}}=\dfrac{0,1}{0,2}=0,5M\)

Câu 2
\(a)PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ b)n_{Mg}=\dfrac{1,2}{24}=0,05mol\\ n_{H_2}=n_{MgCl_2}=n_{Mg}=0,05mol\\ V_{H_2}=0,05.24,79=1,2395l\\ c)n_{HCl}=2n_{Mg}=2.0,05=0,1mol\\ V_{ddHCl}=\dfrac{0,1}{2}=0,05l\\ d)C_{M_{MgCl_2}}=\dfrac{0,05}{0,05}=1M\)

\(a,n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\\ m_{H_2SO_4}=200.19,6\%=39,2\left(g\right)\\ \rightarrow n_{H_2SO_4}=\dfrac{39,2}{98}=0,4\left(mol\right)\)

PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)

bđ        0,3    0,4

pư        0,3        0,3

spư      0           0,1           0,3        0,3

\(\rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)

\(b,m_{dd}=19,5+200-0,3.2=218,9\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}C\%_{ZnSO_4}=\dfrac{0,3.161}{218,9}.100\%=22,06\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,1.98}{218,9}.100\%=4,48\%\end{matrix}\right.\)

Tham Khảo