Câu hỏi của Phạm Gia Linh

Question

1/ Cho a(g) CaCO3 vào 200g dung dịch HCl 10,95% thu được dung dịch A và V(l) khí ở đktc. Để trung hòa dung dịch A cần dùng hết 50 ml dung dịch NaOH 2M.

a) Viết PT phản ứng

b) Tính khối lượng CaCO3 t...

Trần Quốc Lộc · Accepted Answer

Câu 1:

$\text{a)         }pthh:CaCO3+2HCl\rightarrow CaCl_2+CO_2+H_2O\left(1\right)\ NaOH+HCl\rightarrow NaCl+H_2O\left(2\right)$

b) $n_{NaOH}=C_M\cdot V=0,05\cdot2=0,1\left(mol\right)$

Theo $pthh\left(2\right):n_{HCl\left(2\right)}=n_{NaOH}=0,1\left(mol\right)$

$m_{HCl}=\dfrac{m_{d^2HCl}\cdot C\%}{100}=\dfrac{200\cdot10,95}{100}=21,9\left(g\right)\
\Rightarrow n_{HCl}=\dfrac{m}{M}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\
\Rightarrow n_{HCl\left(1\right)}=0,6-0,1=0,5\left(mol\right)$

Theo $pthh\left(1\right):n_{CaCO_3}=\dfrac{1}{2}n_{HCl\left(1\right)}=\dfrac{1}{2}\cdot0,5=0,25\left(mol\right)$

$\Rightarrow m_{CaCO_3}=n\cdot M=0,25\cdot100=25\left(g\right)$

c) Theo $pthh\left(1\right):n_{CO_2}=\dfrac{1}{2}n_{HCl\left(1\right)}=\dfrac{1}{2}\cdot0,5=0,25\left(mol\right)$

$\Rightarrow V_{CO_2}=n\cdot22,4=0,25\cdot22,4=5,6\left(l\right)$

d)  $m_{CO_2}=n\cdot M=0,25\cdot44=11\left(g\right)$

$m_{HCl\left(dư\right)}=n\cdot M=0,1\cdot36,5=3,65\left(g\right)$

$m_{d^2A}=\left(m_{CaCO_3}+m_{d^2HCl}\right)-m_{CO_2}\
=\left(25+200\right)-11=214\left(g\right)$

$\Rightarrow C\%\left(HCl_{dư}\right)=\dfrac{3,65\cdot100}{214}=1,71\%$Câu trả lời: Câu 1:

$\text{a)         }pthh:CaCO3+2HCl\rightarrow CaCl_2+CO_2+H_2O\left(1\right)\ NaOH+HCl\rightarrow NaCl+H_2O\left(2\right)$

b) $n_{NaOH}=C_M\cdot V=0,05\cdot2=0,1\left(mol\right)$

Theo $pthh\left(2\right):n_{HCl\left(2\right)}=n_{NaOH}=0,1\left(mol\right)$

$m_{HCl}=\dfrac{m_{d^2HCl}\cdot C\%}{100}=\dfrac{200\cdot10,95}{100}=21,9\left(g\right)\
\Rightarrow n_{HCl}=\dfrac{m}{M}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\
\Rightarrow n_{HCl\left(1\right)}=0,6-0,1=0,5\left(mol\right)$

Theo $pthh\left(1\right):n_{CaCO_3}=\dfrac{1}{2}n_{HCl\left(1\right)}=\dfrac{1}{2}\cdot0,5=0,25\left(mol\right)$

$\Rightarrow m_{CaCO_3}=n\cdot M=0,25\cdot100=25\left(g\right)$

c) Theo $pthh\left(1\right):n_{CO_2}=\dfrac{1}{2}n_{HCl\left(1\right)}=\dfrac{1}{2}\cdot0,5=0,25\left(mol\right)$

$\Rightarrow V_{CO_2}=n\cdot22,4=0,25\cdot22,4=5,6\left(l\right)$

d)  $m_{CO_2}=n\cdot M=0,25\cdot44=11\left(g\right)$

$m_{HCl\left(dư\right)}=n\cdot M=0,1\cdot36,5=3,65\left(g\right)$

$m_{d^2A}=\left(m_{CaCO_3}+m_{d^2HCl}\right)-m_{CO_2}\
=\left(25+200\right)-11=214\left(g\right)$

$\Rightarrow C\%\left(HCl_{dư}\right)=\dfrac{3,65\cdot100}{214}=1,71\%$

Trần Quốc Lộc · Answer

a) $m_{NaOH}=\dfrac{m_{d^2}\cdot C\%}{100}=\dfrac{200\cdot15}{100}=30\left(g\right)$

$m_{d^2A\text{     sau khi pha thêm      }100\left(g\right)nước}=100+200=300\left(g\right)$

$\Rightarrow C\%\left(NaOH\right)=\dfrac{m_{NaOH}\cdot100}{m_{d^2}}=\dfrac{30\cdot100}{300}=10\%$

b) $m_{NaOH\text{      sau khi cho thêm    }5\left(g\right)NaOH}=30+5=35\left(g\right)$

$m_{d^2B}=5+200=205\left(g\right)$

$\Rightarrow C\%\left(NaOH\right)=\dfrac{m_{NaOH}\cdot100}{m_{d^2}}=\dfrac{35\cdot100}{205}=17,07\%$

c) $C\%\left(NaOH\right)=\dfrac{30\cdot100}{150}=20\%$Câu trả lời: a) $m_{NaOH}=\dfrac{m_{d^2}\cdot C\%}{100}=\dfrac{200\cdot15}{100}=30\left(g\right)$

$m_{d^2A\text{     sau khi pha thêm      }100\left(g\right)nước}=100+200=300\left(g\right)$

$\Rightarrow C\%\left(NaOH\right)=\dfrac{m_{NaOH}\cdot100}{m_{d^2}}=\dfrac{30\cdot100}{300}=10\%$

b) $m_{NaOH\text{      sau khi cho thêm    }5\left(g\right)NaOH}=30+5=35\left(g\right)$

$m_{d^2B}=5+200=205\left(g\right)$

$\Rightarrow C\%\left(NaOH\right)=\dfrac{m_{NaOH}\cdot100}{m_{d^2}}=\dfrac{35\cdot100}{205}=17,07\%$

c) $C\%\left(NaOH\right)=\dfrac{30\cdot100}{150}=20\%$

Bài 40: Dung dịch

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