Question
What volume of dry hydrogen would be produced under standard consitions when 0.250g of magnesium reacts with excess HCl?
Answers
Dr Russ
For this sort of problem always start with a balanced equation,
Mg + HCl -> H2 + MgCl2
which you need to balance.
molar mass for Mg is 24.3 g mole^-1
number of moles (M) of Mg is
M= 0.260 g / 24.3 g mole^-1
Then use the equation to find the number of moles of H2.
Under standard conditios one mole of any gas occupires 22.4 litres
Hence find the volume of gas.
Mg + HCl -> H2 + MgCl2
which you need to balance.
molar mass for Mg is 24.3 g mole^-1
number of moles (M) of Mg is
M= 0.260 g / 24.3 g mole^-1
Then use the equation to find the number of moles of H2.
Under standard conditios one mole of any gas occupires 22.4 litres
Hence find the volume of gas.