What volume of oxygen is produced when 200cm³ of hydrogen peroxide with concentration of 2.5moldm⁻³ fully decomposes? Give your answer to 3s.f. in dm³ and assume reaction takes place at STP.
To find the volume of oxygen produced when the hydrogen peroxide fully decomposes, we need to use the balanced chemical equation for the decomposition of hydrogen peroxide:
2H2O2(l) -> 2H2O(l) + O2(g)
From the equation, we can see that for every 2 moles of hydrogen peroxide, we get 1 mole of oxygen gas. Therefore, we need to calculate the number of moles of hydrogen peroxide in order to find the volume of oxygen produced.
First, let's convert the volume of hydrogen peroxide given from cm³ to dm³. Since 1 dm³ is equal to 1000 cm³, the volume of hydrogen peroxide is:
200 cm³ = 200/1000 dm³ = 0.2 dm³
Next, we calculate the number of moles of hydrogen peroxide using its molarity (concentration) and volume:
Number of moles of hydrogen peroxide = concentration x volume
= 2.5 mol/dm³ x 0.2 dm³
= 0.5 moles
Since the balanced chemical equation shows that 2 moles of hydrogen peroxide produce 1 mole of oxygen gas, we can determine the number of moles of oxygen gas produced:
Number of moles of oxygen gas = 0.5 moles ÷ 2
= 0.25 moles
Finally, we can convert the number of moles of oxygen gas to volume using the ideal gas law at STP (Standard Temperature and Pressure), where 1 mole of any gas occupies a volume of 22.7 dm³:
Volume of oxygen gas = number of moles x molar volume at STP
= 0.25 moles x 22.7 dm³/mole
= 5.675 dm³
Rounding to 3 significant figures, the volume of oxygen gas produced when 200 cm³ of hydrogen peroxide with a concentration of 2.5 mol/dm³ fully decomposes is 5.68 dm³.