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

3 Answers

  • Admire
    Lv 7
    7 years ago
    Favorite Answer

    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

    • Commenter avatarLogin to reply the answers
  • 7 years ago

    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

    • Commenter avatarLogin to reply the answers
  • 7 years ago

    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.

    • Commenter avatarLogin to reply the answers
Still have questions? Get your answers by asking now.