What volume of dry hydrogen would be produced under standard consitions when 0.250g of magnesium reacts with excess HCl?

Answer 1

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.