In the reaction# Mg (s) + 2HCl (aq) -> H_2 (g) + MgCl_2 (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?

1 Answer
Feb 7, 2016

Answer:

# 0.756g#

Explanation:

I am assuming that #6.0 M HCl# is a typo, and that it should be #6.0 molL^-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 litre, and #V# is the volume of the substance in litres.

#n(HCl) = 6.0molL^-1 xx 0.125L = 0.75mol#

Then we find out how many moles of hydrogen gas (#H_2#) are produced. In the formula we see #2HCl#, and #H_2#. This means there is 1 mole of #H_2# for every 2 moles of #HCl# so to find the amount of #H_2# we use:

#n(H_2) = 1/2 xx 0.75mol = 0.375mol#

Now we find the molar mass of the #H_2# 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(H_2) = 0.375mol xx 2.016gmol^-1 = 0.756g#