Ta có: \(m_{HCl}=300.7,3\%=21,9\left(g\right)\Rightarrow n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

___0,3_____0,6_____0,3____0,3 (mol)

a, m_Mg = 0,3.24 = 7,2 (g)

b, Ta có: m _{dd sau pư} = m_Mg + m _{dd HCl} - m_H2 = 7,2 + 300 - 0,3.2 = 306,6 (g)

\(\Rightarrow C\%_{MgCl_2}=\dfrac{0,3.95}{306,6}.100\%\approx9,3\%\)

Bạn tham khảo nhé!

pthh:Mg+2HCl→MgCl2+H2(1)

theo pthh=>\(nMg=\dfrac{1}{2}nHCL=\dfrac{300.7,3\%}{7,3}.\dfrac{1}{2}=\dfrac{3.1}{2}=1,5mol\)

=>mMg=\(1,5.24=36g\)

b, theo pthh(1)\(=>nMgCl2=\dfrac{1}{2}nHCL=1,5mol\)

\(=>mMgCl2=\)\(1,5.95=142,5g\)

\(mdd=\text{ m Mg + mdd HCl - m H2}=36+300-1,5.2=333g\)

\(=>\%mMgCl2=\dfrac{142,5}{333}.100\%=42,8\%\)

1) nZn=13/65=0,2(mol)

PTHH: Zn +  2 HCl -> ZnCl2 + H2

nH2=nZnCl2=nZn=0,2(mol)

nHCl=2.0,2=0,4(mol)

=> mHCl=0,4 x 36,5=14,6(g)

=> mddHCl=(14,6.100)/8=182,5(g)

2) V(H2,đktc)=0,2 x 22,4= 4,48(l)

mZnCl2=0,2.136=27,2(g)

3) mddsau=mZn+mddHCl - mH2= 13+182,5-0,2.2=195,1(g)

4) C%ddZnCl2=(27,2/195,1).100=13,941%

a) $n_{CaCO_3} = 0,15(mol)$
$CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
$n_{HCl} = 2n_{CaCO_3} = 0,3(mol)$
$m_{dd\ HCl} = \dfrac{0,3.36,5}{7,3\%} = 150(gam)$

b)

$n_{CaCl_2} = n_{CO_2} = n_{CaCO_3} =0,15(mol)$
$V_{CO_2} = 0,15.22,4 = 3,36(lít)$
c)

$m_{dd} = 15 + 150 - 0,15.44 = 158,4(gam)$

$C\%_{CaCl_2} = \dfrac{0,15.111}{158,4}.100\% = 10,51\%$

\(n_{HCl}=0,1.0,2=0,02\left(mol\right)\)

Pt : \(2HCl+Ca\left(OH\right)_2\rightarrow CaCl_2+2H_2O\)

       0,02---->0,01---------->0,01

a) Nồng độ mol đề cho rồi mà nhỉ 

b) \(m_{muôi}=m_{CaCl2}=0,01.111=1,11\left(g\right)\)

1) 

a) Từ trái qua phải : 

\(4FeS+7O_2\xrightarrow[]{t^o}2Fe_2O_3+4SO_2\)

\(2SO_2+O_2\xrightarrow[V_2O_5]{t^o}2SO_3\)

\(SO_3+H_2O\rightarrow H_2SO_4\)

\(Cu+2H_2SO_{4\left(đ,n\right)}\rightarrow CuSO_4+H_2O+SO_2\uparrow\)

Bạn xem lại chỗ H₂SO₄ cho ra Cu nhé

b) Từ trái qua phải : 

\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

\(Al_2\left(SO_4\right)_3+3NaOH\rightarrow Al\left(OH\right)_3+3Na_2SO_4\)

\(2Al\left(OH\right)_3\xrightarrow[]{t^o}Al_2O_3+3H_2O\)

2) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

0,2 -->0,6--------->0,2------>0,3

a) \(m_{ddHCl}=\dfrac{\left(0,6.36,5\right)}{7,3\%}.100\%=300\left(g\right)\)

b) \(m_{ddspu}=5,4+300-0,3.2=304,8\left(g\right)\)

\(C\%_{AlCl3}=\dfrac{0,2.133,5}{304,8}.100\%=8,75\%\)

a)

$AgNO_3 + HCl \to AgCl + HNO_3$
Theo PTHH : 

$n_{AgCl} = n_{HCl} = n_{AgNO_3} = \dfrac{340.10\%}{170} =0,2(mol)$

$m_{dd\ HCl} = \dfrac{0,2.36,5}{7,3\%} = 100(gam)$

b)

$m_{AgCl} = 0,2.143,5 = 28,7(gam)$

c)

$m_{dd\ sau\ pư} = 340 + 100 -28,7 = 411,3(gam)$

$n_{HNO_3} = n_{AgNO_3} = 0,2(mol)$
$\Rightarrow C\%_{HNO_3} = \dfrac{0,2.63}{411,3}.100\% = 3,06\%$

200ml = 0,2l

\(n_{Ba\left(OH\right)2}=0,5.0,2=0,1\left(mol\right)\)

Pt : \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O|\)

              1              2             1            2

             0,1           0,2          0,1

a) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)

\(V_{ddHCl}=\dfrac{0,2}{1}=0,2\left(l\right)=200\left(ml\right)\)

b) \(n_{BaCl2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)

⇒ \(m_{BaCl2}=0,1.208=20,8\left(g\right)\)

c) \(V_{ddspu}=0,2+0,2=0,4\left(l\right)\)

\(C_{M_{BaCl2}}=\dfrac{0,1}{0,4}=0,25\left(M\right)\)

 Chúc bạn học tốt

PTHH: \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

Ta có: \(n_{Ba\left(OH\right)_2}=0,2\cdot0,5=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,2\left(mol\right)\\n_{BaCl_2}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{ddHCl}=\dfrac{0,2}{1}=0,2\left(l\right)=200\left(ml\right)\\m_{BaCl_2}=0,1\cdot208=20,8\left(g\right)\\C_{M_{BaCl_2}}=\dfrac{0,1}{0,2+0,2}=0,25\left(M\right)\end{matrix}\right.\)

PƯ: Ba(OH)₂ + Na₂SO₄ = BaSO₄ + 2NaOH (1)
0,08 -----> 0,08 (mol)
Ta có: n_Ba(OH)2= \(\dfrac{40.34.2\%}{100\%.171}\)= 0,08 (mol)
Theo pư (1) có : n_Na2SO4 0,08 (mol)
=> m_Na2SO4 = 0.08.142 = 11,36 (g)
=> m_ddNa2SO4= \(\dfrac{11,36}{14,2}.100\%\)= 80 (g)