Đáp án:
% m Z n = 61 , 9 % ; % m C u = 38 , 1 % \% {m_{Zn}} = 61,9\% ; \% {m_{Cu}} = 38,1\% % m Z n = 6 1 , 9 % ; % m C u = 3 8 , 1 %
m d d H 2 S O 4 = 98 g a m m_{dd H_2SO_4}=98 gam m d d H 2 S O 4 = 9 8 g a m
Giải thích các bước giải:
C u Cu C u H 2 S O 4 H_2SO_4 H 2 S O 4
Phản ứng xảy ra:
Z n + H 2 S O 4 → Z n S O 4 + H 2 Zn + {H_2}S{O_4}\xrightarrow{{}}ZnS{O_4} + {H_2} Z n + H 2 S O 4 Z n S O 4 + H 2
Ta có:
n H 2 = 2 , 24 22 , 4 = 0 , 1 mol = n Z n {n_{{H_2}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol = }}{{\text{n}}_{Zn}} n H 2 = 2 2 , 4 2 , 2 4 = 0 , 1 mol = n Z n
→ m Z n = 0 , 1.65 = 6 , 5 gam \to {m_{Zn}} = 0,1.65 = 6,5{\text{ gam}} → m Z n = 0 , 1 . 6 5 = 6 , 5 gam
→ % m Z n = 6 , 5 10 , 5 = 61 , 9 % → % m C u = 38 , 1 % \to \% {m_{Zn}} = \frac{{6,5}}{{10,5}} = 61,9\% \to \% {m_{Cu}} = 38,1\% → % m Z n = 1 0 , 5 6 , 5 = 6 1 , 9 % → % m C u = 3 8 , 1 %
n H 2 S O 4 = n Z n = 0 , 1 mol {n_{{H_2}S{O_4}}} = {n_{Zn}} = 0,1{\text{ mol}} n H 2 S O 4 = n Z n = 0 , 1 mol
→ m H 2 S O 4 = 0 , 1.98 = 9 , 8 gam \to {m_{{H_2}S{O_4}}} = 0,1.98 = 9,8{\text{ gam}} → m H 2 S O 4 = 0 , 1 . 9 8 = 9 , 8 gam
→ m d d H 2 S O 4 = 9 , 8 10 % = 98 gam \to {m_{dd{\text{ }}{{\text{H}}_2}S{O_4}}} = \frac{{9,8}}{{10\% }} = 98{\text{ gam}} → m d d H 2 S O 4 = 1 0 % 9 , 8 = 9 8 gam
BTKL:
m Z n + m d d   H 2 S O 4 = m d d + m H 2 {m_{Zn}} + {m_{dd\;{{\text{H}}_2}S{O_4}}} = {m_{dd}} + {m_{{H_2}}} m Z n + m d d H 2 S O 4 = m d d + m H 2
→ 0 , 1.65 + 98 = m d d + 0 , 1.2 → m d d = 104 , 3 gam \to 0,1.65 + 98 = {m_{dd}} + 0,1.2 \to {m_{dd}} = 104,3{\text{ gam}} → 0 , 1 . 6 5 + 9 8 = m d d + 0 , 1 . 2 → m d d = 1 0 4 , 3 gam
n Z n S O 4 = n Z n = 0 , 1 → m Z n S O 4 = 0 , 1. ( 65 + 96 ) = 16 , 1 gam {n_{ZnS{O_4}}} = {n_{Zn}} = 0,1 \to {m_{ZnS{O_4}}} = 0,1.(65 + 96) = 16,1{\text{ gam}} n Z n S O 4 = n Z n = 0 , 1 → m Z n S O 4 = 0 , 1 . ( 6 5 + 9 6 ) = 1 6 , 1 gam
→ C % Z n S O 4 = 16 , 1 104 , 3 = 15 , 436 % \to C{\% _{ZnS{O_4}}} = \frac{{16,1}}{{104,3}} = 15,436\% → C % Z n S O 4 = 1 0 4 , 3 1 6 , 1 = 1 5 , 4 3 6 %
