Students in chemistry class have been given the assignment to use flame test emission data to determine the identity of an
1 answer:
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Answer:
- 13.56 g of sodium chloride are theoretically yielded.
- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.
- 0.50 g of sodium nitrate remain when the reaction stops.
- 92.9 % is the percent yield.
Explanation:
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In this case, according to the question, it is possible to set up the following chemical reaction:
Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:
Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:
Therefore, the leftover of sodium nitrate is:
Finally, the percent yield is computed via:
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The correct ratio of the atoms of the two elements in the sample of copper chloride compound would be 1:2.
<h3>Atoms of elements</h3>Copper chloride has a chemical formula, .
Thus, an atom of copper, Cu, combines with 2 atoms of chlorine, Cl.
The 2 chlorine atoms are recipients of an electron each from the copper atom which happens to have 2 electrons in its outermost valence shell.
Therefore, the ratio of copper to chlorine atoms in the compound is 1:2.
More on atoms of elements can be found here: brainly.com/question/15173445
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