Cho 16,8 gam Fe tác dụng với 5,376 lít khí Fl2 ( đo đktc ) sau phản ứng hoàn toàn lấy chất rắn hòa tan vào nước thu được m gam muối. Tính m gam muối

1 câu trả lời

$n_{Fe} = \frac{m_{Fe}}{M_{Fe}} = \frac{16,8}{56} = 0,3 (mol)$

$n_{Fl_{2}} = \frac{V_{Fl_{2}}}{22,4} = \frac{5,376}{22,4} = 0,24 (mol)$

$PTHH:           2Fe + 3Fl_{2} → 2FeFl_{3}$

Ban đầu:       2           3              2

Phản ứng:  0,6  →   0,72    →   0,6

Sau p/ứng:              0,12           0,6

⇒ $Fl_{2}$ dư, $Fe_{}$ hết  

$m_{FeFl_{3}} = n_{FeFl_{3}} . M_{FeFl_{3}} = 0,6 . (56 + 19 . 3) = 67.8 (g)$

Vậy sau phản ứng, ta thu được 67,8 g muối $FeFl_{3}$.

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