5.74 g of mg reacts with hcl to form mgcl and h .calculate the moles of h from 0.0840 of mg and excess hcl
I think you need to do some work first, before I help.
What is the balanced reaction? you have to write the symbols and compounds correctly: Hint, doing that is likely to give you the answer.
PLEASE learn where the caps key is on your computer and USE it.
Second, your question is confusing. Is that two problems or one? I'll ignore the 5.74 g Mg; I don't know what that has to do with the problem. Now, dealing only with the 0.0840 g Mg (by the way, mg is milligrams and not Mg for magnesium).
Mg + 2HCl --> MgCl2 + H2
mols Mg = grams/atomic mass = ?
Now, look at the coefficients in the balanced equation. 1 molMg will produce 1 mol H2; therefore, mols H2 produced will be the same as mols Mg you started with.
Post your work if you need furthere assistance.
To calculate the moles of hydrogen gas (H2), we need to use the stoichiometry of the balanced chemical equation and the given amount of magnesium (Mg) and excess hydrochloric acid (HCl).
First, let's examine the balanced chemical equation for the reaction:
Mg + 2HCl -> MgCl2 + H2
From the equation, we can see that 1 mole of magnesium reacts with 2 moles of hydrochloric acid, producing 1 mole of hydrogen gas.
Given:
Mass of Mg = 0.0840 g
Molar mass of Mg = 24.305 g/mol
First, convert the mass of Mg to moles:
Moles of Mg = Mass of Mg / Molar mass of Mg
= 0.0840 g / 24.305 g/mol
= 0.003456 mol
Next, since magnesium is in excess, the hydrochloric acid is the limiting reactant. This means all the HCl will react, and we can consider the moles of HCl as the moles of hydrogen gas (H2) formed.
Given: Excess HCl
Therefore, the moles of hydrogen gas formed from the given amount of Mg and excess HCl is 0.003456 mol.