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

Answer:

$m_{Mg}=291.6gMg$

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:

$m_{Mg}=5.0\frac{molHCl}{L}*4.8L*\frac{1molMg}{2molHCl}*\frac{24.3gMg}{1molMg}\\m_{Mg}=291.6gMg$

Best regards.

Answer 2

Answer:

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

Explanation:

Molarity of teh HCl solution = 5.0 M

Volume of HCl solution = 4.8 L

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

$5.0 M=\frac{Moles}{4.8 L}$

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:

$\frac{1}{2}\times 24 mol=12 mol$ 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.