How many grams of Mg are needed to react completely with 3.80 L of a 2.50 M HF solution?
Mg + 2HF --> MgF^2 + H^2
Just follow the steps in this example.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the number of grams of Mg needed to react completely with 3.80 L of a 2.50 M HF solution, we need to follow these steps:
Step 1: Write and balance the chemical equation.
The balanced equation for the reaction between Mg and HF is:
Mg + 2HF -> MgF2 + H2
Step 2: Calculate the moles of HF in the solution.
To do this, we use the given concentration and volume of the HF solution:
moles of HF = concentration of HF (M) × volume of HF solution (L)
moles of HF = 2.50 M × 3.80 L
Step 3: Determine the molar ratio between HF and Mg.
From the balanced equation, we can see that the ratio is 1:1. This means that 1 mole of Mg reacts with 1 mole of HF.
Step 4: Calculate the moles of Mg required.
Since the moles of HF and Mg are in a 1:1 ratio, the moles of Mg required will be the same as the moles of HF:
moles of Mg = moles of HF
Step 5: Convert moles of Mg to grams.
To convert moles to grams, we need the molar mass of Mg. The molar mass of Mg is approximately 24.31 g/mol.
grams of Mg = moles of Mg × molar mass of Mg
Now, let's calculate the answer:
moles of HF = 2.50 M × 3.80 L = 9.5 moles
moles of Mg = 9.5 moles
grams of Mg = 9.5 moles × 24.31 g/mol
Therefore, you would need approximately 230.495 grams of Mg to react completely with 3.80 L of a 2.50 M HF solution.