Students heated red mercury(II) oxide until they got a positive splint test for oxygen. They also...
Question:
Students heated red mercury(II) oxide until they got a positive splint test for oxygen. They also noticed that the test tube was coated with silvery mercury metal, which they were inhaling. If a student used 45.5 grams of {eq}HgO {/eq} in this experiment, how many grams of mercury was he inhaling?
{eq}HgO \to Hg + O_2 {/eq} (unbalanced)
Stoichiometry:
Assuming that the reaction is 100% efficient, the amount of product that can be formed from a certain amount of the reactant can be calculated. In this calculation, it provides approximation on how much should one prepare in order to produce a certain amount of product.
Answer and Explanation: 1
The first thing to do is to balance the reaction first.
{eq}\mathbf{2 Hg O \rightarrow 2 Hg + O_2}. {/eq}
From the balanced equation, it can be seen that 2 moles of mercuric oxide reacts to form 2 moles of mercury. This relationship will be used in the calculation.
{eq}\rm Theoretical\ yield\ of\ Hg = (45.5\ g\ HgO)\left(\dfrac{1\ mol\ HgO}{216.59\ g\ HgO}\right)\left(\dfrac{2\ mol\ Hg}{2\ mol\ HgO}\right)(200.59 \ g/mol\ Hg)\\ Theoretical\ yield\ of\ Hg = (0.2101\ mol\ HgO)\left(\dfrac{2\ mol\ Hg}{2\ mol\ HgO}\right)(200.59 \ g/mol\ Hg)\\ Theoretical\ yield\ of\ Hg = (0.2101\ mol\ Hg)(200.59 \ g/mol\ Hg) = \boxed{\mathbf{42.1\ g\ Hg}} {/eq}
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