Đặt:
nZnCO3= x mol
nZnO= y mol
ZnCO3 -to-> ZnO + CO2
x___________x_____x
mhh= 125x + 81y = 8 g (1)
nZnO= 6.24/81= 52/675 mol
<=> x + y = 52/675 (2)
Giải (1) và (2) :
x= 0.04
y= 1/27
mZnCO3= 0.04*125=5g
mZnO= 1/27*81= 3g
%ZnCO3= 62.5%
%ZnO= 37.5%
b) 2KOH + CO2 --> K2CO3 + H2O
0.08______0.04
mKOH= 0.08*56=4.48g
a)mCO2= 8-6.24=1.76
nCO2= 1.76/44=0.04 mol
nZnCO3=nCO2=0.04 mol
mZnCO3=0.04*125=5g
%mZnCO3=5*100/8=62.5%
%ZnO= 37.5%
$n_{ZnO}=\dfrac{6,24}{81}= \dfrac{52}{675} mol \\PTHH : \\ZnCO_3 \overset{t^o}\to ZnO + CO2↑(1) \\a.Gọi\ n_{ZnCO_3}=x;n_{ZnO}=y(x,y>0) \\Ta\ có : \\m_{hh}= 125x + 81y = 8 g \\n_{ZnO}=x + y = \dfrac{52}{675} mol$
$\text{Ta có hpt :}$
$\left\{\begin{matrix} 125x + 81y = 8 & \\ x + y = \dfrac{52}{675} & \end{matrix}\right.⇔\left\{\begin{matrix} x=0,04 & \\ y=\dfrac{1}{27} & \end{matrix}\right. \\⇒\%m_{ZnCO_3}=\dfrac{0,04.125}{8}.100\%= 62,5\% \\\%m_{ZnO}=100\%-62,5\%= 37,5% \\b) 2KOH + CO_2 \to K_2CO_3 + H_2O(2)$
$\text{Theo pt (1) : }$
$n_{CO_2}=n_{ZnCO_3}=0,4mol$
$\text{Theo pt (2) :}$
$n_{KOH}=2.n_{CO_2}=2.0,04=0,08mol$
$⇒m_{KOH}=0,08.56=4,48g$