Question

cho 8.125 gam kẽm tác dụng với 250 gam dung dịch axit hcl 7.3 %

a, tính thẻ tích khí hidroo thu được ở dktc .

b, tính c% chất có trong dung dịch sau phản ứng

Answer 1

Ta co pthh

Zn + 2HCl \(\rightarrow\) ZnCl2 + H2

Theo de bai ta co

nZn= \(\dfrac{8,125}{65}=0,125mol\)

mct=mHCl=\(\dfrac{mdd.C\%}{100\%}=\dfrac{250.7,3\%}{100\%}=18,25g\)

\(\Rightarrow nHCl=\dfrac{18,25}{36,5}=0,5mol\)

Theo pthh ta co

nZn=\(\dfrac{0,125}{1}mol< nHCl=\dfrac{0,5}{2}mol\)

-> so mol cua HCl du (tinh theo so mol cua Zn )

a,Theo pthh

nH2=nZn=0,125 mol

\(\Rightarrow VH2_{\left(dktc\right)}=0,125.22,4=2,8\left(l\right)\)

b,Theo pthh

nZnCl2=nZn=0,125 mol

nHCl=2nZn=2.0,125 = 0,25 mol

\(\Rightarrow\) mZnCl2 = 0,125 .136=17 g

mHCl(du) = (0,5 - 0,25 ).36,5=9,125 g

\(\Rightarrow\) mddZnCl2 = mZn + mddHCl - mH2 = 8,125 + 250 -(0,125.2) = 257,875 g

\(\Rightarrow\) C% \(_{ddZnCl2}=\dfrac{17}{257,875}.100\%\approx6,59\%\)

C%\(_{ddHCl\left(du\right)}=\dfrac{9,125}{257,875}.100\%\approx3,539\%\)

Answer 2

a, PTHH: Zn + 2HCl -> ZnCl₂ + H₂

Ta có: \(n_{Zn}=\dfrac{8,125}{65}=0,125\left(mol\right)\\ m_{HCl}=\dfrac{250.7,3}{100}=18,25\left(g\right)\\ =>n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)

=> \(\dfrac{0,125}{1}< \dfrac{0,5}{2}\)

=> Zn hết, HCl dư tính theo Zn.

=> \(n_{H_2}=n_{Zn}=0,125\left(mol\right)\\ =>V_{H_2\left(đktc\right)}=0,125.22,4=2,8\left(l\right)\)

b, Chất có trong dd sau khi phản ứng kết thúc là HCl và ZnCl₂.
Ta có: \(m_{H_2}=0,125.2=0,25\left(g\right)\\ n_{ZnCl_2}=n_{Zn}=0,125\left(mol\right)\\ =>m_{ZnCl_2}=0,125.136=17\left(g\right)\\ n_{HCl\left(f.ứ\right)}=2.0,125=0,25\left(mol\right)\\ =>n_{HCl\left(dư\right)}=0,5-0,25=0,25\left(mol\right)\\ =>m_{HCl\left(dư\right)}=0,25.36,5=9,125\left(g\right)\)

Ta có: \(m_{ddsau-phản-ứng}=8,125+250-0,25=258,875\left(g\right)\)

Nồng đô phần trăm các chất thu dc sau khi phản ứng kết thúc:

\(C\%_{ddZnCl_2}=\dfrac{17}{258,875}.100\approx6,567\%\)

\(C\%_{ddHCl\left(dư\right)}=\dfrac{9,125}{258,875}.100\approx3,525\%\)

Answer 3

a) PTHH: Zn + 2HCl --> ZnCl₂ + H₂

Ta có : n_Zn = \(\dfrac{8,125}{65}\) = 0,125 mol

m_HCl = 250. 7,3% = 18,25 (g)

=> n_HCl = \(\dfrac{18,25}{36,5}\) = 0,5 mol

Vì \(\dfrac{0,125}{1}\) < \(\dfrac{0,5}{2}\) => HCl dư

Cứ 1 mol Zn --> 1 mol H₂

0,125 mol --> 0,125 mol

=> \(V_{H_2}\) = 0,125 . 22,4 = 2,8 l

b) Áp dụng ĐLBTKL ta có:

m_Zn + m_HCl = m_{dd sau p/ứ} + m_H2

=> m_{dd sau p/ứ} = 8,125 + 250 - 0,125.2 = 257,875 (g)

=> \(C\%_{ZnCl_2}\) ₌ \(\dfrac{0,125\times136}{257,875}\times100\%\) = 6,59%

=> C%_{HCl dư} = \(\dfrac{\left(0,5-0,125\times2\right)\times36,5}{257,875}\times100\%\) = 3,54%

Answer 4

\(Zn+2HCl->ZnCl_2+H_2\)

n_Zn= 8,125/65=0,125mol

m_HCl =250.7,3%=18,25 gam

=> n_HCl =18,25/36,5=0,5 mol

\(\dfrac{0,125}{1}< \dfrac{0,5}{2}=>HCl,dư\)

=> n_H2= n_Zn = n_ZnCl2 = 0,125 mol

=> V_H2 =0,125.22,4=2,8 lít

mdd_{sau-phản-ứng} =8,125+250-0,125.2=207,875 gam

C%(ZnCl₂) = \(\dfrac{0,125.136}{207,875}.100\%=8,178\%\)

nHCl,dư = 0,5-0,125.20,25 mol => mHCl_dư =0,25.36,5=9,125gam

C%(HCl,dư) = \(\dfrac{9,125}{207,875}.100\%=4,39\%\)