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

Since the Mg is in excess, therefore HCl will be fully consumed in the reaction.

The first step is to find the amount of HCl in mol

Let  N (HCl) = amount of HCl in mol

N (HCl) = (6 mol HCL/L solution) *( 125 mL ) * (1 L/1000 mL) = 0.75 mol of HCl

Through stoichiometry

N (H2) = 0.75 mol HCl * (1 mol H2/ 2 mol of HCl)

N(H2) = 0.375 mol H2

 Since we are asked for the number of grams of H2 (mass), we multiply this with the molar mass of hydrogen

M (H2) = 0.375 mol H2 ( 2 g H2 / 1 mol H2)

M (H2) = 0.75 g H2 

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

i think the answer is terrae but not 100 percent sure. ▓▬▬ §