Start by writing down the balanced chemical equation that describes this single replacement reaction
Now, notice that the two reactants react in a #1:color(red)(2)# mole ratio. This tells you that the reaction will always consume twice as many moles of hydrochloric acid than moles of zinc metal. As you know, a solution's molarity tells you how many moles of solute, which in your case will be hydrochloric acid, #"HCl"#, are present per liter of solution. The hydrochloric acid solution is said to have a molarity of #"3.05 mol L"^(-1)#. This means that you get #3.05# moles of hydrochloric acid for every #"1 L"# of solution. Keep this in mind for later. You know that the reaction must consume #"25.0 g"# of zinc metal. You can convert this mass to moles by using zinc's molar mass
In order for this many moles of zinc to react, you'd need
Since you know that #"1 L"# of your hydrochloric acid solution contains #3.05# moles of hydrochloric acid, ti follows that this many moles would be present in
Rounded to three sig figs and expressed in milliliters, the answer will be
Note that you have
SIDE NOTE The hydrogen gas produced by the reaction will bubble out of solution. With an understanding of the ideal gas laws, it is now possible to apply these principles to chemical stoichiometry problems. For example, zinc metal and hydrochloric acid (hydrogen chloride dissolved in water) react to form zinc (II) chloride and hydrogen gas according to the equation shown below: 2 HCl (aq) + Zn (s) → ZnCl2 (aq) + H2 (g)
A sample of pure zinc with a mass of 5.98 g is reacted with excess hydrochloric acid and the (dry) hydrogen gas is collected at 25.0 ˚C and 742 mm Hg. What volume of hydrogen gas would be produced?
This is a “single state” problem, so we can solve it using the ideal gas law, PV = nRT. In order to find the volume of hydrogen gas (V), we need to know the number of moles of hydrogen that will be produced by the reaction. Our stoichiometry is simply one mole of hydrogen per mole of zinc, so we need to know the number of moles of zinc that are present in 5.98 grams of zinc metal. The temperature is given in centigrade, so we need to convert into Kelvin, and we also need to convert mm Hg into atm. Conversions: \[25.0\; C+273=298\; K \nonumber \] \[(742\; mm\; Hg)\times \left ( \frac{1\; atm}{760\; mm\; Hg} \right )=0.976\; atm \nonumber \] \[(5.98\; g\; Zn)\times \left ( \frac{1.00\; mol}{65.39\; g\; Zn} \right )=0.0915\; mol \nonumber \] Substituting: \[PV=nRT \nonumber \] \[(0.976\; atm)\times V=(0.0915\; mol)(0.0821\; L\; atm\; mol^{-1}K^{-1})(298\; K) \nonumber \] \[V=\frac{(0.0915\; mol)(0.0821\; L\; atm\; mol^{-1}K^{-1})(298\; K)}{(0.976\; atm)}=2.29\; L \nonumber \] We can also use the fact that one mole of a gas occupies 22.414 L at STP in order to calculate the number of moles of a gas that is produced in a reaction. For example, the organic molecule ethane (CH3CH3) reacts with oxygen to give carbon dioxide and water according to the equation shown below: 2 CH3CH3 (g) + 7 O2 (g) → 4 CO2 (g) + 6 H2O (g)
An unknown mass of ethane is allowed to react with excess oxygen and the carbon dioxide produced is separated and collected. The carbon dioxide collected is found to occupy 11.23 L at STP; what mass of ethane was in the original sample?
Because the volume of carbon dioxide is measured at STP, the observed value can be converted directly into moles of carbon dioxide by dividing by 22.414 L mol–1. Once moles of carbon dioxide are known, the stoichiometry of the problem can be used to directly give moles of ethane (molar mass 30.07 g mol-1), which leads directly to the mass of ethane in the sample. \[(11.23\; L\; CO_{2})\times \left ( \frac{1\; mol}{22.414\; L} \right )=0.501\; mol\; CO_{2} \nonumber \] Reaction stoichiometry: \[(0.501\; mol\; CO_{2})\times \left ( \frac{2\; mol\; CH_{3}CH_{3}}{4\; mol\; CO_{2}} \right )=0.250\; mol\; CH_{3}CH_{3} \nonumber \] The ideal gas laws allow a quantitative analysis of whole spectrum of chemical reactions. When you are approaching these problems, remember to first decide on the class of the problem:
Once you have isolated your approach ideal gas law problems are no more complex that the stoichiometry problems we have addressed in earlier chapters.
2 NaN3 (s) → 2 Na (s) + 3 N2 (g) What mass of sodium azide is necessary to produce the required volume of nitrogen at 25 ˚C and 1 atm?
2 Fe2O3(s) + 3 C (s) → 4 Fe (s) + 3 CO2 (g)
Zn (s) + 2 HCl (aq) → ZnCl2 (aq) + H2 (g) |