Answer:
The volume of hydrogen gas produced will be approximately 50.7 liters under STP.
Explanation:
Relative atomic mass data from a modern periodic table:
Magnesium is a reactive metal. It reacts with hydrochloric acid to produce
- Hydrogen gas [tex]\rm H_2[/tex], and
- Magnesium chloride, which is a salt.
The chemical equation will be something like
[tex]\rm ?\;Mg\;(s) + ?\;HCl \;(aq)\to ?\;H_2 \;(g)+ [\text{Formula of the Salt}][/tex],
where the coefficients and the formula of the salt are to be found.
To determine the number of moles of [tex]\rm H_2[/tex] that will be produced, first find the formula of the salt, magnesium chloride.
Magnesium is a group 2 metal. The oxidation state of magnesium in compounds tends to be +2.
On the other hand, the charge on each chloride ion is -1. Each magnesium ion needs to pair up with two chloride ions for the charge to balance in the salt, magnesium chloride. The formula for the salt will be [tex]\rm MgCl_2[/tex].
[tex]\rm ?\;Mg\;(s) + ?\;HCl\;(aq) \to ?\;H_2 \;(g)+ ?\;MgCl_2\;(aq)[/tex].
Balance the equation. [tex]\rm MgCl_2[/tex] contains the largest number of atoms among all species in this reaction. Start by setting its coefficient to 1.
[tex]\rm ?\;Mg\;(s) + ?\;HCl\;(aq) \to ?\;H_2 \;(g)+ {\bf 1\;MgCl_2}\;(aq)[/tex].
The number of [tex]\rm Mg[/tex] and [tex]\rm Cl[/tex] atoms shall be the same on both sides. Therefore
[tex]\rm {\bf 1\;Mg}\;(s) + {\bf 2\;HCl}\;(aq) \to ?\;H_2 \;(g)+ {1\;\underset{\wedge}{Mg}\underset{\wedge}{Cl_2}}\;(aq)[/tex].
The number of [tex]\rm H[/tex] atoms shall also conserve. Hence the equation:
[tex]\rm {1\;Mg}\;(s) + {2\;\underset{\wedge}{H}Cl}\;(aq) \to {\bf 1\;H_2 \;(g)}+ {1\;MgCl_2}\;(aq)[/tex].
How many moles of HCl are available?
[tex]M(\rm HCl) = 1.008 + 35.45 = 36.458\;g\cdot mol^{-1}[/tex].
[tex]\displaystyle n({\rm HCl}) = \frac{m(\text{HCl})}{M(\text{HCl})} = \rm \frac{165.0\;g}{36.458\;g\cdot mol^{-1}} = 4.52576\;mol[/tex].
How many moles of Hydrogen gas will be produced?
Refer to the balanced chemical equation, the coefficient in front of [tex]\rm HCl[/tex] is 2 while the coefficient in front of [tex]\rm H_2[/tex] is 1. In other words, it will take two moles of [tex]\rm HCl[/tex] to produce one mole of [tex]\rm H_2[/tex]. [tex]\rm 4.52576\;mol[/tex] of [tex]\rm HCl[/tex] will produce only one half as much [tex]\rm H_2[/tex].
Alternatively, consider the ratio between the coefficient in front of [tex]\rm H_2[/tex] and [tex]\rm HCl[/tex] is:
[tex]\displaystyle \frac{n(\text{H}_2)}{n(\text{HCl})} = \frac{1}{2}[/tex].
[tex]\displaystyle n(\text{H}_2) = n(\text{HCl})\cdot \frac{n(\text{H}_2)}{n(\text{HCl})} = \frac{1}{2}\;n(\text{HCl}) = \rm \frac{1}{2}\times 4.52576\;mol = 2.26288\;mol[/tex].
What will be the volume of that many hydrogen gas?
One mole of an ideal gas occupies a volume of 22.4 liters under STP (where the pressure is 1 atm.) On certain textbook where STP is defined as [tex]\rm 1.00\times 10^{5}\;Pa[/tex], that volume will be 22.7 liters.
[tex]V(\text{H}_2) = \rm 2.26288\;mol\times 22.4\;L\cdot mol^{-1} = 50.69\; L[/tex], or
[tex]V(\text{H}_2) = \rm 2.26288\;mol\times 22.7\;L\cdot mol^{-1} = 51.37\; L[/tex].
The value "165.0 grams" from the question comes with four significant figures. Keep more significant figures than that in calculations. Round the final result to four significant figures.
