Question

Chỉ em b bài1 với bài 2 đi ạ em cảm ơn

Answer 1

Câu 1:

Ta có: \(m_{dd}=\dfrac{25}{50\%}=50\left(g\right)\) \(\Rightarrow m_{H_2O}=m_{dd}-m_{đường}=25\left(g\right)\)

Answer 2

Câu 2: 

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

Ta có: \(n_{HCl}=\dfrac{200\cdot7,3\%}{36,5}=0,4\left(mol\right)\)

\(\Rightarrow n_{CaCl_2}=0,2\left(mol\right)=n_{CO_2}=n_{CaCO_3}\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaCO_3}=0,2\cdot100=20\left(g\right)\\m_{CaCl_2}=0,2\cdot111=22,2\left(g\right)\\m_{CO_2}=0,2\cdot44=8,8\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd\left(sau.p/ứ\right)}=m_{CaCO_3}+m_{ddHCl}-m_{CO_2}=211,2\left(g\right)\)

\(\Rightarrow C\%_{CaCl_2}=\dfrac{22,2}{211,2}\cdot100\%\approx10,51\%\)

 

Answer 3

Bài 1:

a) mdd= \(\dfrac{25.100}{50}\) = 50 (g)

b) mH₂O= 50-25=25 (g)

Bài 2:

a) CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

b) m_HCl = \(\dfrac{200.7,3}{100}=14,6\) (g)

n_HCl = \(\dfrac{14,6}{36,5}=0,4\) (mol)

nCaCO₃ = nCO₂ = nCaCl₂ = \(\dfrac{1}{2}\)nHCl= \(\dfrac{0,4}{2}\) = 0,2 (mol)

mdd = mddHCl + mCaCO₃ - mCO₂

    = 200 + 0,2.100 - 0,2.44 = 211,2 g

C%CaCl₂= \(\dfrac{0,2.111}{211,2}.100\%\)= 10,51%