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ĐỂ TRUNG HÒA  11.2G DUNG DỊCH KOH 25% THÌ CẦN DÙNG BAO NHIÊU GAM DUNG DỊCH H2SO4 4.9%

Hoàng Thị Ngọc Anh · Accepted Answer

$2KOH\left(0,05\right)+H_2SO_4\left(0,025\right)\rightarrow K_2SO_4+2H_2O$

$m_{KOH}=11,2.25\%=2,8g$

$\Rightarrow n_{KOH}=\dfrac{2,8}{56}=0,05mol$

$m_{H_2SO_4}=0,025.98=2,45g$

$\Rightarrow m_{ddH_2SO_4}=\dfrac{100.2,45}{4,9}=50g.$Câu trả lời: $2KOH\left(0,05\right)+H_2SO_4\left(0,025\right)\rightarrow K_2SO_4+2H_2O$

$m_{KOH}=11,2.25\%=2,8g$

$\Rightarrow n_{KOH}=\dfrac{2,8}{56}=0,05mol$

$m_{H_2SO_4}=0,025.98=2,45g$

$\Rightarrow m_{ddH_2SO_4}=\dfrac{100.2,45}{4,9}=50g.$

Hong Ra On · Answer

$m_{KOH}=\dfrac{11,2.25}{100}=2,8\left(g\right)$

$n_{KOH}=\dfrac{2,8}{56}=0,05\left(mol\right)$

PTHH: $2KOH+H_2SO_4-->K_2SO_4+2H_2$

$m_{H_2SO_4}=\dfrac{1}{2}.0,05.98=2,45\left(g\right)$

$\Rightarrow m_{ddH_2SO_4}=\dfrac{2,45.100}{4,9}=50\left(g\right)$Câu trả lời: $m_{KOH}=\dfrac{11,2.25}{100}=2,8\left(g\right)$

$n_{KOH}=\dfrac{2,8}{56}=0,05\left(mol\right)$

PTHH: $2KOH+H_2SO_4-->K_2SO_4+2H_2$

$m_{H_2SO_4}=\dfrac{1}{2}.0,05.98=2,45\left(g\right)$

$\Rightarrow m_{ddH_2SO_4}=\dfrac{2,45.100}{4,9}=50\left(g\right)$

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