How many elements are in “C2H6O”
2 answers:
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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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Answer:
15.2 mL
Explanation:
First we calculate the volume of the copper
when this volume of copper is added to the water, the water level will rise by the level of the volume of copper. So the final volume is:
14.6 mL + 0.5580 mL = 15.158 mL
Since the measuring cylinder is graduated to one decimal place , we can round this up to 15.2 mL