# equation calculations

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#1
Calculate the mass of NaOH that reacts completely with 0.1mol / dm³ of 250 cm³ H2SO4 solution. .
Calculate the volume of 0.15mol / d * m ^ 3 NaOH solution to neutralize 0.1mol/dm³ of 250cm³ of H2SO4solution
Last edited by Anonymous🥱; 11 months ago
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11 months ago
#2
I’m not sure what H.50 and how do you have area (m^2) of a solution are you sure you copied the question out right.

Normally on these you work out the moles of one thing. Then use the balanced equation/the molar ratios to work out the moles of the thing your looking for. With part A you would multiply by Mr to find the mass. With part B you would divide by the concentration (0.15).
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#3
Oh sorry it's cm³
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11 months ago
#4
(Original post by Anonymous🥱)
Oh sorry it's cm³
Any idea what H.50 is?
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11 months ago
#5
Is it 0.5 moldm^3 H+
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#6
Calculate the mass of NaOH that reacts completely with 0.1mol / dm³ of 250 cm³ H2SO4 solution. .
Calculate the volume of 0.15mol / d * m ^ 3 NaOH solution to neutralize 0.1mol/dm³ of 250cm³ of H2SO4solution
0
#7
Calculate the mass of NaOH that reacts completely with 0.1mol / dm³ of 250 cm³ H2SO4 solution. .
Calculate the volume of 0.15mol / d * m ^ 3 NaOH solution to neutralize 0.1mol/dm³ of 250cm³ of H2SO4solution
0
11 months ago
#8
(Original post by Anonymous🥱)
Calculate the mass of NaOH that reacts completely with 0.1mol / dm³ of 250 cm³ H2SO4 solution. .
Calculate the volume of 0.15mol / d * m ^ 3 NaOH solution to neutralize 0.1mol/dm³ of 250cm³ of H2SO4solution
Ok so for part A work out the moles of H2SO4. The reaction is 2NaOH + H2SO4 -> Na2SO4 +2H2O so double the moles of H2SO4 to find moles of NaOH. Then finally multiply by the Mr.

For part B you’ve already found the moles on NaOH in part A. Just divide by the concentration.
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