If Chelcy and Allyson reacted 2.55 grams of Mg with excess hydrochloric acid (HCl), how many grams of MgCl2 will they theoretically produce? Mg(s) + HCl (aq) → H2 (g) + MgCl2 (aq)

?g MgCl2

18.5

To determine the grams of MgCl2 produced, we need to use stoichiometry to find the molar ratio between Mg and MgCl2.

Step 1: Calculate the molar mass of MgCl2.
The molar mass of magnesium chloride (MgCl2) can be calculated by adding the atomic masses of each element:
Molar mass of Mg = 24.31 g/mol
Molar mass of Cl = 35.45 g/mol (since there are 2 Cl atoms in MgCl2)

Molar mass of MgCl2 = (24.31 g/mol) + 2*(35.45 g/mol) = 95.21 g/mol

Step 2: Calculate the number of moles of Mg reacted.
Given mass of Mg = 2.55 grams
Molar mass of Mg = 24.31 g/mol

Number of moles of Mg = Mass of Mg / Molar mass of Mg
Number of moles of Mg = 2.55 g / 24.31 g/mol

Step 3: Use stoichiometry to determine the moles of MgCl2 produced.
From the balanced equation, the stoichiometric ratio between Mg and MgCl2 is 1:1. This means that for every one mole of Mg reacted, one mole of MgCl2 is produced.

Number of moles of MgCl2 = Number of moles of Mg (from step 2)

Step 4: Convert moles of MgCl2 to grams.
Grams of MgCl2 = Number of moles of MgCl2 * Molar mass of MgCl2
Grams of MgCl2 = Number of moles of Mg (from step 3) * Molar mass of MgCl2 (from step 1)

Substituting the calculated values:
Grams of MgCl2 = (2.55 g / 24.31 g/mol) * 95.21 g/mol

Calculating this expression will give you the number of grams of MgCl2 produced.

To find the theoretical yield of MgCl2, you need to determine the limiting reactant in the chemical equation. The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed.

To determine the limiting reactant, you need to compare the amount of reactants given in the problem. In this case, Chelcy and Allyson reacted 2.55 grams of Mg.

Now, you need to convert the mass of Mg to moles. To do this, divide the mass of Mg by its molar mass. The molar mass of Mg is 24.31 g/mol.

2.55 g Mg x (1 mol Mg / 24.31 g Mg) = 0.105 mol Mg

According to the balanced equation, the stoichiometric ratio between Mg and MgCl2 is 1:1. This means that for each mole of Mg reacted, 1 mole of MgCl2 is produced.

Therefore, the number of moles of MgCl2 produced is also 0.105 mol.

Finally, to convert moles of MgCl2 to grams, multiply the number of moles by the molar mass of MgCl2. The molar mass of MgCl2 is 95.21 g/mol.

0.105 mol MgCl2 x (95.21 g MgCl2 / 1 mol MgCl2) ≈ 10.02 grams of MgCl2

So, Chelcy and Allyson would theoretically produce approximately 10.02 grams of MgCl2.

Mg + 2HCl ==> MgCl2 + H2

mols Mg = grams/atomic mass.
Using the coefficients in the balanced equation, convert mols Mg to mols mgCl2 (looks like 1:1 to me).
Now convert mols MgCl2 to grams. g = mols x molar mass.