Question

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

Answer 1

Answer:

0.756g

Step-by-step explanation:

I am assuming that 6.0MHCl is a typo, and that it should be 6.0molL−1HCl , 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)=6.0molL−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×2.016gmol−1=0.756g

Answer 2

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