Đáp án:
Giải thích các bước giải:
$a/$
có : $m_{HCl} = 200.14,6\% = 29,2(gam)$
$⇒ n_{HCl} = \dfrac{29,2}{36,5} = 0,8(mol)$
$NaOH + HCl → NaCl + H_2O$
Theo PTHH: $n_{NaOH} = n_{HCl} = 0,8(mol)$
$⇒ m_{\text{dung dịch NaOH}} = \dfrac{0,8.40}{8\%} = 400(gam)$
$b/$
Sau phản ứng :
$m_{dd} = m_{\text{dd HCl}} + m_{\text{dd NaOH}} = 200 + 400 = 600(gam)$
$n_{NaCl} = n_{HCl} = 0,8(mol)$
$⇒ C\%_{NaCl} = \dfrac{0,8.58,5}{600}.100\% = 7,8\%$
a) mHCl=$\frac{mđHCl×C%ddHCl}{100}$ =$\frac{200×14,6}{100}$=29,2
nHCl=$\frac{29,2}{36,5}$ =0,8(mol)
PTHH: NaOH + HCl→NaCl+ H2O
TBR : 0,8 ←0,8→ 0,8 (mol)
mNaOH=0,8×40=32(g)
mdd NaOH=$\frac{32×100}{8}$ =400(g)
b)mNaCl=0,8×58,5=46,8(g)
mdd sau=mddHCl+mNaOH=200+32=232
C%dd NaCl=$\frac{mNaCl×100}{mdd sau }$ =$\frac{46,8×100}{232}$ =20,17%
Cho x + y = 3 Tính A = x^2+2xy+y^2-4x-4y+1
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