In the reaction mg (s) + 2hcl (aq) h2 (g) + mgcl2 (aq), how many moles of hydrogen gas will be produced from 75.0 milliliters of a 1.0 m hcl in an excess of mg?

HCl is limiting reactant, Mg in excess therefore:

number of moles HCl:

75.0 mL in liters: 75.0 / 1000 => 0.075 L

Molarity HCl = 1.0 M

n = M x V

n = 1.0 x 0.075 

n = 0.075 moles of HCl

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

2 moles HCl   1 mole H2
0.075 moles of HCl ( moles H2)

moles H2 = 0.075 x 1 / 2

moles H2 = 0.075 / 2

= 0.0375 moles of H2

hope this helps!

The answer is c fish