Calculate the volume of hydrogen gas that is produced by the reaction of 0.52 grams of solid magnesium with excess hydrochloride acid (HCl). This is a single replacement reaction.

In this given case, he volume of hydrogen gas produced = 0.03969 L

Given:

0.52 grams - solid magnesium

Solution:

moles of Mg

=

= 0.216 x 10⁻³  

Reaction -

Mg + 2HCl -----------------> MgCl2 + H2  

  1mol                                                     1mol  

0.216 x 10⁻³                                     0.216 x 10⁻³ mol  

moles of H2

= 0.216 x 10⁻³  

T  = 273 K

P =1 atm

From idea gas P V = n R T

Placing given and known values in ideal gas

1 x V = 0.216 x 10⁻³ x 0.0821 x 273  

V = 0.048 L  

Thus, the volume of hydrogen gas produced = 0.03969 L

Learn more:

link

0.493L

Explanation:

Let us write a balanced equation for the reaction. This is illustrated below:

Mg + 2HCl —> MgCl2 + H2

Next, let us convert 0.52g of Mg to mole. This is illustrated below:

Molar Mass of Mg = 24g/mol

Mass of Mg = 0.52g

Number of mole of Mg = Mass /Molar Mass = 0.52/24 = 0.022mole

From the equation,

1mole of Mg produced 1mole of H2.

Therefore, 0.022mole of Mg will always produce 0.022mole of H2.

1mole of H2 occupies 22.4L at stp.

Therefore, 0.022mole of H2 will occupy = 0.022 x 22.4 = 0.493L

Al escribirla, he tenido cuatro objetivos presentes. En primer lugar, re- trasé su redacción y publicación hasta que estuve bastante seguro de no in-.

270 páginas