Đáp án:
`a, 24gam`
`b, 260,869ml`
Giải thích các bước giải:
`a,`
`K_2O + H_2O -> 2KOH + H_2`
`n_{KOH}=0,4.1=0,4(mol)`
theo PTHH:
`n_{K_2O}= 1/2 .n_{KOH}=1/2 .0,4=0,2(mol)`
`-> m_{K_2O}=0,2.94=18,8(gam)`
`-> m_{CuO}=42,8-18,8=24(gam)`
`b,`
`CuO + 2HCl -> CuCl_2 + H_2O`
`n_{CuO}=(24)/(80)=0,3(mol)`
Theo pthh:
`n_{HCl}=2.n_{CuO}=2.0,3=0,6(mol)`
`C1:`
`-> m_{HCl}=0,6.36,5=21,9(gam)`
`-> m_{dd HCl}= 21,9 : 7,3%=300(gam)`
`-> V_{dd HCl}= m/D = (300)/(1,15)=260,869(ml)`
`C2:`
`C_{M_{dd HCl}}=(10.1,15)/(36,5).7,3=2,3(M)`
`-> V_{dd HCl}=n/(C_M)= (0,6)/(2,3)=0,260869(lít)=260,869ml`