What volume will be released when 5 g of magnesium interacts with 8 g of hydrochloric acid (HCL) solution.

Given:
m (Mg) = 5 g
m (НCl) = 8 g

Find: V (MgCl 2) -?

Decision:
0.208 mol 0.216 mol x mol
Mg + 2НCl = MgCl2 + Н 2
1 mol 2 mol 1 mol 1 mol

1) find the amount of substance (N) magnesium.
To do this, we divide the mass of the thing by its molar mass:
N (Mg) = m / M = 5g / 24 g / mol = 0.208 mol

2) Now we will find the number of HCl things:
N (НCl) = 8g / 37 g / mol = 0.216 mol

3) Let us find out which thing is in excess and which is in short supply. Let’s find the proportionality coefficient:
Mg 0.208 mol / 1 mol = 0.208-excess
HCl 0.216 mol / 2 mol = 0.108-deficiency

4) we solve the problem by the lack. Let’s denote by x what is unknown and solve the proportion:
X = 0.216×1 / 2 = 0.108

N (MgCl2) = 0.108 mol

5) Find the volume of MgCl2. To do this, multiply the number of things by the constant Avagadro:
V (MgCl 2) = 0.108 mol x 22.4 l / mol = 2.4192 l

if you need to find the volume of gas, then everything is the same up to point number 4
Next, you need to designate for X the number of things hydrogen and inscribe X mol on top
4) we solve the problem by the lack. Let’s denote by x what is unknown and solve the proportion:
X = 0.216×1 = 0.108 (but the values ​​matched only in this case)
2
5) Find the volume of H2. To do this, multiply the number of things by the constant Avagadro:
V (H2) = 0.108 mol x 22.4 L / mol = 2.4192 L

Answer: 2.4192 L