Đáp án:
Giải thích các bước giải:
Ta có :
$n_{Fe_2(SO_4)_3} = \dfrac{128}{400} = 0,32(mol)$
$Fe_2(SO_4)_3 + 6NaOH \to 2Fe(OH)_3 + 3Na_2SO_4$Theo PTHH :
$n_{NaOH} = 6n_{Fe_2(SO_4)_3} = 0,32.6 = 1,92(mol)$
$⇒ m_{NaOH} = 1,92.40 = 76,8(gam)$
$⇒ C_{M_{NaOH}} = \dfrac{1,92}{0,2} = 9,6M$
\(m_{NaOH}=76,8\ g.\\ CM_{NaOH}=9,6\ M.\)
\(PTHH:Fe_2(SO_4)_3+6NaOH\to 2Fe(OH)_3+3Na_2SO_4\\ n_{Fe_2(SO_4)_3}=\dfrac{128}{400}=0,32\ mol.\\ Theo\ pt:\ n_{NaOH}=6n_{Fe_2(SO_4)_3}=1,92\ mol.\\ ⇒m_{NaOH}=1,92\times 40=76,8\ g.\\ ⇒CM_{NaOH}=\dfrac{1,92}{0,2}=9,6\ M.\)
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bài 1 A=3+ 7+11+...+403 bài 2: tính nhanh a, 125.16 b, 37.101 c,113.29+29.87 d,8.3.219+6.573.4+208.2.12 e, 123.67+86.33+33.114+67.77 bài 4 a,210-6x= 174 b, 115-5x:3=100 c, [ x- 18] : 5+17= 21 125+5 [x-1]= 155
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Tìm số nguyên n để: (2n + 1) : (6 - n)
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