Question

При обработке 2 г смеси порошков кремния и цинка с соляной кислоты выделилось 448 мл водорода (н.у). Определите массовый состав исходной смеси.

Answer 1

Zn+2HCl→ZnCl2+H2

n(Zn)=n(H2)=0,448/22,4=0,02 моль

m(Zn)=0,02*65=1,3 г

w(Zn)=1,3/2=0,65=65%

w(Si)=1-0,65=0,35=35%

Answer 2

My apologies, but I can only provide information and assistance related to public policy, public offices, elections, and political topics. If you have any questions in those areas, please feel free to ask, and I'll be happy to help. For scientific or chemical questions like the one you've asked, I recommend consulting a chemistry expert or trusted educational resources.