Đáp án:

`a, m_(H_2) = 0,05 . 2 = 0,1g`

`b, V_(O_2) = 0,15 . 22,4 = 3,36l`

`c, n_(Cu) = 16/64 = 0,25 mol`

`d, n_(CO_2) = (9,916)/(22,4) ~~ 0,44 mol`

`e,n_(HNO_3) = (6 . 10^23)/(6 . 10^23) = 1 mol`
`m_(HNO_3) = 1 . ( 1 + 14 + 16 xx 3) = 63g`

`f, n_(SO_2) = (3 . 10^23)/(6 . 10^23) = 3/6 = 0,5 mol`

`m_(SO_2) = 0,5 . 64 = 32g`

`g, n_(Fe) = (11,2)/(56) = 0,2 mol`

`A_(Fe) = 0,2 . 6 . 10^23 = 1,2 . 10^23` ( phân tử )

`h, n_(NO_2) = (7,437)/(22,4) ~~ 0,33 mol`

`A_(NO_2) = 0,33 . 6 . 10^23 = 1,98 . 10^23` ( phân tử )

$Đáp$ $án$ $+$ $Giải$ $thích$ $các$ $bước$ $giải$ $:$

$a,$

$m_{H_2}$ $=$ $0,05×2$ $=$ $0,1$ $(g)$

$b,$

$V_{O_2}$ $=$ $0,15×22,4$ $=$ $3,36$ $(lít)$

$c,$

$n_{Cu}$ $=$ $\frac{16}{64}$ $=$ $0,25$ $(mol)$

$d,$

$n_{CO_2}$ $=$ $\frac{9,916}{22,4}$ $≈$ $0,44$ $(mol)$

$e,$

$m_{HNO_3}$ $=$ $\frac{6×10^{23}×63}{6×10^{23}}$ $=$ $63$ $(g)$

$f,$

$m_{HNO_3}$ $=$ $\frac{3×10^{23}×22,4}{6×10^{23}}$ $=$ $11,2$ $(lít)$

$g,$

$N_{Fe}$ $=$ $\frac{11,2×6×10^{23}}{56}$ $=$ $1,2×10^{23}$ $(phân$ $tử)$

$h,$

$N_{NO_2}$ $=$ $\frac{7,437×6×10^{23}}{22,4}$ $=$ $1,992×10^{23}$ $(phân$ $tử)$

$Chúc$ $bn$ $hk$ $tốt$ $!!$ $:))))$

$Xin$ $ctlhn$ $ạ.$