Question

How many grams of magnesium metal will react completely with 4.8 liters of 5.0 M HCl? Show all of the work needed to solve this problem.

Answer 1

`m_(Mg)=291.6gMg`

Step-by-step explanation:

Hello,

At first the balanced chemical reaction between magnesium metal and hydrochloric acid is shown below:

`Mg+2HCl-->MgCl_2+H_2`

Magnesium grams are computed via the following stoichiometric procedure considering a 1/2 by mole relationship between magnesium and hydrochloric acid in the chemical reaction:

`m_(Mg)=5.0(molHCl)/(L)*4.8L*(1molMg)/(2molHCl)*(24.3gMg)/(1molMg)\\m_(Mg)=291.6gMg`

Best regards.

Answer 2

291.72 grams of magnesium metal will react completely with 4.8 liters of 5.0 M HCl.

Step-by-step explanation:

Molarity of teh HCl solution = 5.0 M

Volume of HCl solution = 4.8 L

`Molarity=\frac{\text{Number of moles Moles}}{\text{Volume of the solution (L)}}`

`5.0 M=(Moles)/(4.8 L)`

Moles of HCl = 24 moles

`Mg+2HCL\rightarrow MgCl_2+H_2`

According to reaction , 2 moles of HCl reacts with 1 mole magnesium.

Then 24 moles of HCl will react with:

`(1)/(2)* 24 mol=12 mol` of magnesium

Mass of 12 mol of magnesium =24.31 g/mol × 12 mol = 291.72 g

291.72 grams of magnesium metal will react completely with 4.8 liters of 5.0 M HCl.