chemistry......?

Question

magnesium was reacted with an excess of dilute hydrochloric acid and the hydrogen gas produced collected in a eudiometer. aThe volume of hydrogen in eudiometer was corrected to conditions of STP. If 94.1 millileter of hydrogen was produced, how much magnesium reacted in this experiment?

i'm stuck...and how can i practice with these kinds of problems/? i need to know it for final in 3 weeks so i have time to learn

Answer 1

In any stochiometry question, your first step is always to write a balanced equation:

Mg(s) + 2HCl(aq) --> MgCl2(aq)+ H2(g)

94,1 mL of H2 was produced and you are at STP. An ideal gas at STP occupies a volume of 22.4 L per mole.

94.1 mL = 0.0941 L
0.0941/22.4 = 0.0042 moles of H2

From the balanced equation, we see that the mole ratio of Mg to H2 is 1:1, so 0.0042 moles of Mg reacted. Your question does not specify whether your answer should be in moles or grams. If your answer is supposed to be in grams then convert from moles to grams:
atomic mass of Mg = 24.3 g/mol

0.0042 moles x 24.3 g/mol = 0.102 g Mg

Answer 2

Repeat after me:
Mass or Volume to Moles by reaction to Moles to mass or volume.

Now a reaction"
Mg + 2HCl -> H2 + MgCl2
We know how much H2 we have by volume. Divide by 22400 mL H2/mole to get moles
The equation tells us that each mole of H2 form takes out 1 g-atom of Mg.
So take the moles of H2, multiply by 1 to g-atoms of Mg
Multiply g-atoms by atomic weight Mg to get answer.

It's important you recognize this pathway. If you get it down cold, you can organize the way you answer any of these stoichiometry problems.

Answer 3

the balanced equation for this reaction

Mg + 2HCl --> MgCl2 + H2

here is the general equation.
(GIVEN / 1) x (MOL GIVEN / WEIGHT GIVEN) x ( MOL RATIO WANTED / MOL RATIO GIVEN) x (WEIGHT WANTED / WEIGHT GIVEN)

in this case, 1 mol of gas at STP is 22.4 L. so

( .0941L H2 / 1) x ( 1 mol H2 / 22.4 L H2) ( 1 Mol Mg / 1 Mol H2) x (24.3 g Mg / 1 Mol Mg)

once you have the equation written, just multiply everything on top and divide it by everything on the bottom and you should get

.1021 g Mg

or

102.1 mg Mg

Answer 4

Mg + 2HCl---> MgCl2 + H2

volume of H2 = 94.1mL =0.0942L
molar volume of gas at stp = 22.4L/mol
moles of H2 at stp = vol of H2/mol voume
= 0.0942L/22.4
=0.004205mol
mol ratio of Mg:H2 = 1:2
moles of Mg reacted =0.004205/2 = 0.002103mol
mass of Mg reacted = 0.002103mol x Ar of Mg
= 0.002103mol x 24.30g/mol
=0.05110g