Question

cho 4,8g Mg vào 36,5g dd HCl 20%. Tính C% dd thu được

Answer 1

Mg + 2HCl \(\rightarrow\) MgCl₂ + H₂

n_Mg = 4,8_/24 = 0,2(mol)

m_HCl = \(\dfrac{m_{dd}.C\%}{100\%}=\dfrac{36,5.20\%}{100\%}=7,3\left(g\right)\)

=> n_HCl = 7,3_/36,5 =0,2(mol)

Lập tỉ lệ :

\(\dfrac{n_{Mg\left(ĐB\right)}}{n_{Mg\left(PT\right)}}=\dfrac{0,2}{1}=0,2\) > \(\dfrac{n_{HCl\left(ĐB\right)}}{n_{HCl\left(PT\right)}}=\dfrac{0,2}{2}=0,1\)

=> Sau phản ứng : Mg dư , HCl hết

=> dd thu được là dd MgCl₂

Theo PT => n_H2 = 1_/2 . n_HCl = 1_/2 . 0,2 = 0,1(mol)

=> m_H2 = 0,1 . 2 = 0,2 (g)

Theo ĐLBTKL : m_{dd sau pứ} = m_Mg + m_{dd HCl} - m_H2 =4,8+ 36,5 - 0,2 =41,1(g)

Theo PT => n_MgCl2 = 1_/2 . n_HCl = 1_/2 . 0,2 =0,1(mol)

=> m_MgCl2 = 0,1 . 95 = 9,5(g)

=> C%_{dd thu được =} 9,5_/41,1 . 100% =23,11%

Answer 2

\(Mg+2HCl(0,2)--->MgCl_2(0,1)+H_2(0,1)\)

\(nMg=0,2(mol)\)

\(mHCl=\dfrac{20.36,5}{100}=7,3\left(g\right)\)

\(\Rightarrow nHCl=0,2\left(mol\right)\)

So sánh \(\dfrac{nMg}{1}=0,2>\dfrac{nHCl}{2}=0,1\)

=> Mg còn dư, chọn nHCl để tính

Theo PTHH: \(nMgCl_2=0,1\left(mol\right)\)

\(\Rightarrow mMgCl_2=9,5\left(g\right)\)

\(nH_2=0,1\left(mol\right)\)

\(\Rightarrow mH_2=0,2\left(g\right)\)

\(m dd sau =4,8+36,5-0,2=41,1(g)\)

\(\Rightarrow C\%_{MgCl_2}=\dfrac{9,5}{41,1}.100\%=23,11\%\)

Answer 3

Ta co pthh

Mg + 2HCl \(\rightarrow\) MgCl2 + H2

Theo de bai ta co

nMg=\(\dfrac{4,8}{24}=0,2mol\)

mHCl=\(\dfrac{mdd.C\%}{100\%}=\dfrac{36,5.20\%}{100\%}=7,3g\Rightarrow nHCl=\dfrac{7,3}{36,5}=0,2mol\)

Theo pthh

nMg=\(\dfrac{0,2}{1}mol>nHCl=\dfrac{0,2}{1}mol\)

\(\Rightarrow\)nMg du ( tinh theo nHCl)

Theo pthh

nMgCl2=\(\dfrac{1}{2}nHCl=\dfrac{1}{2}.0,2=0,1mol\)

\(\Rightarrow\) mMgCl2=0,1.95=9,5 g

Theo pthh

nH2=\(\dfrac{1}{2}nMg=\dfrac{1}{2}.0,2=0,1mol\)

\(\Rightarrow\) mH2=0,1.2=0,2 g

\(\Rightarrow\) mdd= mct + mddHCl - mchatkhi=4,8+36,5-0,2=41,1 g

\(\Rightarrow\) Nong do % chat thu duoc la

C%=\(\dfrac{mct}{mdd}.100\%=\dfrac{9,5}{41,1}.100\%=23,1\%\)