a) mct = \(\dfrac{150.13,1}{100}\) = 19,65 g
b)
nZn = \(\dfrac{6,5}{65}\) = 0,1 mol
Zn + 2HCl -> ZnCl2 + H2
0,1 ->0,1 ->0,1
=>mZnCl2 = 0,1 . 136 = 13,6 g
C% = \(\dfrac{13,6}{6,5+500-0,1.2}\).100% = 2,68%
a) Theo bài ra, ta có:
mFe= 150 . 13,1 % : 100% = 19,65 (g)
b) PTHH: Zn + 2HCl → ZnCl2 + H2↑
Theo bài ra, ta có:
nZnCl2=nZn= 0,1 (mol)
⇔ mZnCl2 = n.M=0,1 . 136 = 13,6 (g)
Theo PTHH, ta có:
nH2= nZn=0,1 (mol)
⇔ mH2= n.M= 0,1 . 2 = 0,,2 (mol)
Ta có: mdd ZnCl2 = mZn+ mHCl- mH2
= 6,5 + 500 - 0,2
= 506,3 (g)
⇔ C% = mct/mdd=13,6 . 100% / 506,3 ≃ 2,69%
\(a)m_{ct}=\dfrac{150.13,1}{100}=19,65\left(g\right)\)
b) \(n_{Zn}=0,1\left(mol\right)\)
pt:
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(0,1\rightarrow0,1\rightarrow0,1\)
\(\Rightarrow m_{ZnCl_2}=136.0,1=13,6\left(g\right)\)
\(C\%=\dfrac{13,6}{6,5+500.2-0,1}.100\%=2,68\%\)
