Question

Cho 13 gam Zn tác dụng với HCl 7,5% a. Tính khối lượng dd HCl. b. Tính C% dd sau phản ứng.

Answer 1

Sửa đề : 7.3%

\(n_{Zn}=\dfrac{13}{65}=0.2\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(0.2........0.4..........0.2.........0.2\)

\(m_{HCl}=0.4\cdot36.5=14.6\left(g\right)\)

\(m_{dd_{HCl}}=\dfrac{14.6\cdot100}{7.3}=200\left(g\right)\)

\(m_{ZnCl_2}=0.2\cdot136=27.2\left(g\right)\)

\(m_{\text{dung dịch sau phản ứng}}=13+200-0.2\cdot2=212.6\left(g\right)\)

\(C\%_{ZnCl_2}=\dfrac{27.2}{212.6}\cdot100\%=12.79\%\)

Answer 2

nZn = 0,2 mol

a, PTHH : Zn + 2HCl -> ZnCl2 + H2

- Theo PTHH : nHCl = 2nZn = 0,4mol

=> mHCl = 14,6g

=> mddHCl = \(\dfrac{584}{3}\)g

b, - Theo PTHH : nZnCl2 = nZn = 0,2 mol

=> mZnCl2 = 27,2g

Mà mdd = mZn + mdd - mH2 = \(\dfrac{3109}{15}\)g

=> C%ZnCl2 = ~13,12%

Answer 3

a) n Zn = 13/65  = 0,2(mol)

Zn + 2HCl → ZnCl2 + H2

0,2....0,4.........0,2........0,2..............(mol)

m dd HCl = 0,4.36,5/7,5% = 194,67(gam)

b)

Sau pư, mdd = m Zn + mdd HCl - m H2 = 13 + 194,67 - 0,2.2 = 207,27(gam)

C% ZnCl2 = 0,2.136/207,27  .100% = 13,12%