Đáp án:

m =32,4g

mddH2SO4 = 49g

Giải thích các bước giải:

a) MgCO3 + H2SO4 → MgSO4 + H2O +CO2 ↑

MgSO4 + 2NaOH → Mg(OH)2 + Na2SO4

$Mg{(OH)_2}\buildrel {to} \over

\longrightarrow MgO + {H_2}O$

b) nCO2 = 2,24 : 22,4 = 0,1mol

nMgCO3 = nCO2 = 0,1 mol

nMgO = 12:40=0,3mol

nMgSO4 = nMgO - nMgCO3 = 0,3 - 0,1 = 0,2mol

m = mMgCO3 + mMgSO4

= 0,1 .84+0,2.120=32,4g

nH2SO4 = nCO2 = 0,1 mol

mH2SO4 = 0,1.98=9,8g

mddH2SO4 = 9,8:20.100=49g

chúc bạn học tốt

a, \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

\(FeCl_3+3KOH\rightarrow3KCl+Fe\left(OH\right)_{3\downarrow}\)

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

Theo PT: \(n_{HCl}=6n_{Fe_2O_3}=0,6\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\)

b, \(n_{Fe\left(OH\right)_3}=n_{FeCl_3}=2n_{Fe_2O_3}=0,2\left(mol\right)\)

\(\Rightarrow m_{Fe\left(OH\right)_3}=0,2.107=21,4\left(g\right)\)

\(n_{KOH}=3n_{FeCl_3}=0,6\left(mol\right)\)

\(\Rightarrow C_{M_{KOH}}=\dfrac{0,6}{0,2}=3\left(M\right)\)

\(a.MgO+2HCl\rightarrow MgCl_2+H_2O\)

\(b.n_{MgO}=\dfrac{16}{40}=0,04mol\)

\(\rightarrow n_{HCl}=0,04.2=0,08mol\)

\(C_{M_{HCl}}=\dfrac{0,08}{0,15}=0,53M\)

\(c.m_{MgCl_2}=0,04.95=3,8g\)

\(n_{H_2SO_4}=\dfrac{200\cdot19.6\%}{98}=0.4\left(mol\right)\)

\(SO_3+H_2O\rightarrow H_2SO_4\)

\(0.4......................0.4\)

\(m_{SO_3}=0.4\cdot80=32\left(g\right)\)

\(b.\)

\(n_{H_2SO_4}=\dfrac{80\cdot19.6\%}{98}=0.16\left(mol\right)\)

\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

\(0.16..........0.16..............0.16\)

\(m_{MgO}=0.16\cdot40=6.4\left(g\right)\)

\(m_{\text{dung dịch sau phản ứng}}=6.4+80=86.4\left(g\right)\)

\(C\%MgSO_4=\dfrac{0.16\cdot120}{86.4}\cdot100\%=22.22\%\)

a)

$SO_3 + H_2O \to H_2SO_4$

n SO3 = n H2SO4 = 200.19,6%/98 = 0,4(mol)

=> m = 0,4.80 = 32(gam)

b)

$MgO + H_2SO_4 \to MgSO_4 + H_2O$

n MgSO4 = n MgO = n H2SO4 = 80.19,6%/98 = 0,16(mol)

=> m MgO = 0,16.40 = 6,4(gam)

Sau pư, m dd =  6,4 + 80 = 86,4(gam)

=> C% MgSO4 = 0,16.120/86,4   .100% = 22,22%

nCO2 = 0,448 / 22,4 = 0,02 (mol)

2CH3COOH + CaCO3 -> (CH3COO)2Ca + CO2 + H2O

0,04 0,02 0,02

mCH3COOH = 0,04 * 60 = 2,4 (gam)

-> C%CH3COOH = 2,4 / 40 * 100% = 6%

mCaCO3 = 0,02 * 100 = 2 (gam)

PTHH: CaCO₃ + 2HCl \(\rightarrow\) CaCl₂ + H₂O + CO₂\(\uparrow\) (1)

Theo pt: .... 1 ......... 2 ............ 1 ......... 1 ....... 1 .... (mol)

Theo đề: .. 0,25 .... 0,5 ........................................... (mol)

PTHH: 2HCl_dư + MgO \(\rightarrow\) MgCl₂ + H₂O (2)

Theo pt: ... 2 .......... 1 ............ 1 .......... 1 .. (mol)

Theo đề: .. 0,5 ..... 0,25 ............................. (mol)

\(n_{MgO}=\dfrac{m}{M}=\dfrac{10}{40}=0,25\left(mol\right)\)

\(m_{HCl_{dư}}=n.M=0,5.36,5=18,25\left(g\right)\)

\(m_{HCl\left(1\right)}=m_{HCl_{bđ}}-m_{HCl_{dư}}=36,5-18,25=18,25\left(g\right)\)

\(n_{HCl\left(1\right)}=\dfrac{m}{M}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)

\(m_{CaCO_3}=n.M=0,25.100=25\left(g\right)\)

Vậy m = 25g

Đặt : 

nFe = x mol 

nMgO = y mol 

mX = 56x + 40y = 13.6 (g) (1) 

Fe +  2HCl => FeCl2 + H2 

x____________x

MgO + 2HCl => MgCl2 + H2O 

y______________y

mM = mFeCl2 + mMgCl2 = 127x + 95y = 31.7 (2) 

(1) , (2) : 

x = 0.1 

y = 0.2 

%Fe = 5.6/13.6 * 100% = 41.17%

%MgO = 58.82%

nKOH = 0.1 * 0.2 = 0.02 (mol) 

KOH + HCl => KCl + H2O 

0.02____0.02 

nHCl (pư) = 2nFe + 2nMgO = 0.1*2 + 0.2*2 = 0.6 (mol) 

nHCl = 0.02 + 0.6 = 0.62 (mol) 

VddHCl = 0.62/0.5 = 1.24 (M) 

nCaCO3 = 45/100 = 0,45 (mol)

PTHH: CaCO3 + 2HCl -> CO2 + H2O + CaCl2 

Mol:  0,45 ---> 0,9 ---> 0,45 ---> 0,45 ---> 0,45 

Số phân tử CO2 = H2O = CaCl2: 0,45 . 6.10^23 = 2,7.10^23 (phân tử)

mCO2 = 0,45 . 44 = 19,8 (g)

mH2O = 0,45 . 18 = 8,1 (g)

mCaCl2 = 49,95 (g)

mHCl = 0,9 . 36,5 = 32,85 (g)

Số phân tử HCl: 0,9 . 6.10^23 = 5,4 10^23 (phân tử)

Đáp án : A

X + NaOH : COOH + NaOH à COONa + H₂O

=> n_COOH (23 – 1) = (n_Ala + 2n_Glutamic ).22 = 30,8 (*)

X + HCl : NH₂ + HCl à NH₃Cl

=> 36,5.n_NH2 = 36,5.(n_Ala + n_Glutamic) = 36,5

=> n_Ala  =0,6 ; n_Glutamic = 0,4 mol

=> m = 112,2g