In the reaction mg (s) + 2hcl (aq) →mgcl2 (aq) + h2 (g) + MgCl2 (aq) how many grams of hydrogen gas will be produced from 125.0 milliliters of a 6.0 M HCI in an excess of Mg

The reaction

Mg (s) + 2HCl (aq) = MgCl2 (aq) + H2 (g) + MgCl2

Moles of HCl used will be;

= 0.125 l * 6.0 M

= 0.75 moles

The mole ratio of HCl to H2 is 2:1

Therefore; moles of Hydrogen is 0.75/2 = 0.375 mole

Molecular mass of hydrogen is 2 g

Hence; the mass of hydrogen = 2 * 0.375 = 0.75 g

0.756g

Explanation:

I am assuming that 6.0MHCl is a typo, and that it should be 6.0mol L - 1 HCl, since that makes sense in the equation.

First we have to find the amount of HCl in the solution. We use the formula n=cV where n is the amount of substance in moles, c is the concentration of the solution in moles per liter, and V is the volume of the substance in liters.

n (HCl) = 6.0mol L - 1 * 0.125L=0.75mol

Then we find out how many moles of hydrogen gas (H2) are produced. In the formula we see 2HCl, and H2. This means there is 1 mole of H2 for every 2 moles of HCl so to find the amount of H2 we use:

n (H2) = 12 * 0.75mol=0.375mol

Now we find the molar mass of the H2 molecules, by adding together the atomic weights of the constituent molecules. In this case: 1.008+1.008=2.016. Then we use the formula m=nM where m is the mass of the substance in grams, and M is the molar mass of the substance in grams per mole.

m (H2) = 0.375mol*2.016gmo l - 1 = 0.756 g