How many grams of Mg are needed to react completely with 3.80 L of a 2.50 M HF solution?
Mg + 2HF MgF2 + H2
Here is a worked example of a stoichiometry problem.
http://www.jiskha.com/science/chemistry/stoichiometry.html
115?
To answer this question, we can use the balanced chemical equation and the concept of stoichiometry.
First, let's calculate the number of moles of HF present in the solution using the given concentration and volume. The formula to calculate the number of moles is:
moles = concentration (M) × volume (L)
moles of HF = 2.50 M × 3.80 L = 9.50 moles
From the balanced equation, we can see that the ratio between Mg and HF is 1:2. This means that for every 1 mole of Mg, we need 2 moles of HF.
So, the number of moles of Mg needed to react completely can be found by dividing the number of moles of HF by the stoichiometric ratio:
moles of Mg = moles of HF / stoichiometric ratio (HF to Mg)
moles of Mg = 9.50 moles / 2 = 4.75 moles
Finally, we can calculate the mass of Mg needed using its molar mass. The molar mass of Mg is approximately 24.31 g/mol.
mass of Mg = moles of Mg × molar mass of Mg
mass of Mg = 4.75 moles × 24.31 g/mol = 115.53 grams
Therefore, approximately 115.53 grams of Mg are needed to react completely with 3.80 L of a 2.50 M HF solution.