the 50 ml of hcl acid contained 0.05mol Hcl. show by calculation why the ballon with .6g Mg inflated to about twice the size of the balloon with 0.3g Mg.
my balanced equation: Mg+2HCl -> H2
[yeah they asked to write one] this is a lab question.
2007-04-02 14:29:08 · 2 answers · asked by Stephanie 3 in Science & Mathematics ➔ Chemistry
Mg + 2HCl -> H2 is not a balanced equation. Your magnesium and chlorine have disappeared! The correct balanced equation is Mg + 2HCl -> H2 + MgCl2.
n(HCl) = 0.05 mol.
So n(Mg) required to completely react with the HCL is 0.05 / 2 = 0.025 mol.
m(Mg) = 24.3 (0.025) = 0.61g.
In the balloon with 0.6g of Mg, just about all the HCl was able to react and form hydrogen gas. In the balloon with 0.3g of Mg, only half the HCl was able to react, and so only half as much hydrogen was produced.
2007-04-02 14:40:34 · answer #1 · answered by Scarlet Manuka 7 · 1⤊ 0⤋
Your "balanced equation" should include MgCl2.
From your equation, for each g-atom (or fraction thereof) of Mg, 1 mole (or fraction thereof) of H2 is produced, AS LONG AS YOU HAVE ENOUGH HCl AVAILABLE. Since Mg has a Atomic Weight of about 24, 0.6 grams corresponds to 0.6/24 moles or 0.025 g-atoms of Mg. Since there is twice as many moles of HCl available, the above is true.
2007-04-02 21:39:07 · answer #2 · answered by cattbarf 7 · 0⤊ 1⤋