1.

m_Na2CO3.10H2O=286.0,35=100,1(g)

Ta có:

n_Na2CO3=n_Na2CO3.10H2O=0,35(mol)

m_Na2CO3=106.0,35=37,1(g)

C% dd Na2CO3=\(\dfrac{37,1}{100,1+234,9}.100\%=11,07\%\)

b;

V_H2O=234,9.1=234,9(ml)

V_{H2O trong Na2CO3.10H2O}=3,5.18.1=63(ml)

C_M Na₂CO₃=\(\dfrac{0,35}{0,063+0,2349}=1,16M\)

2.

K₂O + H₂O \(\rightarrow\)2KOH

n_K2O=\(\dfrac{47}{94}\)=0,5(mol)

Theo PTHH ta có:

\(\dfrac{1}{2}\)n_K2O=n_KOH=0,25(mol)

m_KOH=0,25.56=14(g)

m_{KOH trong dd}=m.\(\dfrac{7,93}{100}=0,0793m\)

Ta có:

\(\dfrac{14+0,0793m}{47+m}.100\%=21\%\)

=>m=31,6(g)

Câu 3 chắc là thiếu đề

Câu 1: Theo đề bài, ta có:\(n_{Na2CO3}=0,35\left(mol\right)\)

\(\Rightarrow m_{Na2CO3}=106.0,35=37,1\left(g\right)\)

a) \(C\%_{ddNa2CO3}=\dfrac{37,1}{0,35.286+234,9}.100\%\approx11,07\%\)

b) Ta có: \(\left\{{}\begin{matrix}V_{H2O\left(1\right)}=234,9.1=234,9\left(ml\right)=0,2349\left(l\right)\\V_{H2O\left(2\right)}=0,35.10.18=63\left(ml\right)=0,063\left(l\right)\end{matrix}\right.\)

\(C_{M\left(ddNa2CO3\right)}=\dfrac{0,35}{0,2349+0,063}\approx1,174\left(M\right)\)

Câu 2: \(n_{K2O}=\dfrac{47}{94}=0,5\left(mol\right)\)

PTHH: \(K_2O+H_2O\rightarrow2KOH\)

pư............0,5.........0,5.............1 (mol)

\(\Rightarrow\left\{{}\begin{matrix}m_{KOH\left(1\right)}=56.1=56\left(g\right)\\m_{KOH\left(2\right)}=7,93\%.m\left(g\right)\end{matrix}\right.\)

Theo đề bài, ta có: \(\dfrac{56+7,93\%m}{47+m}.100\%=21\%\)

\(\Rightarrow m\approx352,95\left(g\right)\)

https://hoc24.vn/cau-hoi/cho-400-g-dung-dich-naoh-30-tac-dung-vua-het-voi-100-g-dung-dich-hcl-tinh-a-nong-do-muoi-thu-duoc-sau-phan-ungb-tinh-nong-do-axit-hcl-bie.7974814552205

Tui trả lời rùi nghen

a)

$Mg + H_2SO_4 \to MgSO_4 + H-2$

b) $n_{H_2SO_4} = n_{Mg} = \dfrac{4,8}{24} = 0,2(mol)$
$C\%_{H_2SO_4} = \dfrac{0,2.98}{200}.100\% = 9,8\%$

$n_{H_2} = n_{Mg} = 0,2(mol)$

$\Rightarrow m_{dd\ A} = 4,8 + 200 - 0,2.2 = 204,4(gam)$
$C\%_{MgSO_4} = \dfrac{0,2.120}{204,4}.100\% = 11,7\%$

c) $V_{H_2} = 0,2.22,4 = 4,48(lít)$

Bạn ơi , câu 1 sao dd NaCl lại tác dụng với NaCl v ?

a) \(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: 2HCl + CaCO₃ → CaCl₂ + CO₂ + H₂O

Mol:      0,4                         0,2        0,2

b) \(C\%_{ddHCl}=\dfrac{0,4.36,5.100\%}{100}=14,6\%\)

c) \(m_{CaCl_2}=0,2.101=20,2\left(g\right)\)