# what volume of in liters, of H2 at STP is released when 8.45g of Mg reacts?

Mg(s)+2HCl(aq) =MgCl2(aq)+H2(g). i'm looking for the formula to get this going

Relevance

Check that equation is balanced. Fine.

1mol of Mg=1mol of H2

Number of moles of Mg=Mass/Molar mass

=8.45/24 =0.352mol

This is the same as the number of moles of H2, 0.352mol

At STP, 1mol=22.4L

0.352mol= 0.352x22.4=7.89L

• Calculate moles of Mg present in 8.45 g

8.45 g x 1 mol/24 g = 0.35 moles Mg

Write the balanced equation: Mg(s)+2HCl(aq) =MgCl2(aq)+H2(g)

Notice that 1 mole Mg gives 1 mole H2 gas.

You have 0.35 moles Mg, so you will get 0.35 moles H2 gas

At STP 1 mole of gas = 22.4 liters, thus 0.35 moles = (0.35)(22.4) = 7.9 liters

• X=atomic mass of Mg. X gram of Mg release 2 gram of H2.therefore 8.45 gram of Mg release 2/X multiply with 8.45 .then we get gram of H2 released.then we should find mole .for that we should divide by 2(molecular mass of H2).then we should multiply with 22.4 .then we get litre of H2 released.