Answer :
Here's how to do it:
Balanced equation first:
Mg + HCl = H2 + MgCl2 unbalanced
Mg + 2 HCl = H2 = MgCl balanced
Therefore 1 mole Mg reacts with 2 moles Hcl.
50g Mg = ? moles (a bit over 2; you work it out)
75 g HCl = ? moles (also a bit over 2; you work it out)
BUT, you need twice the moles HCl; therefore it is the Mg that is in excess. (you can now work out how many moles are in excess, and therefore how much mg is left over).
So, 2 moles HCl produce 1 mole H2(g)
therefore, the amount of H2 produced is half the number of moles of HCl
At STP, there are X litres per mole of gas (look it up - about 22 from memory)
Therefore, knowing the moles of H2, you can calculate the volume
Balanced equation first:
Mg + HCl = H2 + MgCl2 unbalanced
Mg + 2 HCl = H2 = MgCl balanced
Therefore 1 mole Mg reacts with 2 moles Hcl.
50g Mg = ? moles (a bit over 2; you work it out)
75 g HCl = ? moles (also a bit over 2; you work it out)
BUT, you need twice the moles HCl; therefore it is the Mg that is in excess. (you can now work out how many moles are in excess, and therefore how much mg is left over).
So, 2 moles HCl produce 1 mole H2(g)
therefore, the amount of H2 produced is half the number of moles of HCl
At STP, there are X litres per mole of gas (look it up - about 22 from memory)
Therefore, knowing the moles of H2, you can calculate the volume