Câu 3:
PTHH:
$2Zn+O_2\buildrel{{t^o}}\over\to 2ZnO$
$2Cu+O_2\buildrel{{t^o}}\over\to 2CuO$
Bảo toàn khối lượng:
$m_{O_2}=m_{\text{oxit}}-m_{KL}=a+8-a=8(g)$
$\Rightarrow n_{O_2}=\dfrac{8}{32}=0,25(mol)$
$2KMnO_4\buildrel{{t^o}}\over\to K_2MnO_4+MnO_2+O_2$
$\Rightarrow n_{KMnO_4}=2n_{O_2}=0,25.2=0,5(mol)$
$\Rightarrow m_{KMnO_4}=158.0,5=79g$
Câu 4:
$n_{O_2}=\dfrac{1.1,92}{0,082(20+273)}=0,08(mol)$
Hao hụt $10\%$ nên cần điều chế $0,08+0,08.10\%=0,088(mol)$
$2KMnO_4\buildrel{{t^o}}\over\to K_2MnO_4+MnO_2+O_2$
$\Rightarrow n_{KMnO_4}=2n_{O_2}=0,176(mol)$
$m_{KMnO_4}=158.0,176=27,808g$