In one experiment, 12.0 g of PCl5 was slowly added to 15.0 g of H2O according to the following balanced chemical equation: Pcl5(s)+4H2O(l) ---> H3PO4 (aq) + 5HCl (g) The molar masses for each compounds in the equation are as follows; PCl5: 208.224 g/mol H2O: 18.015 g/mol H3PO4: 97.994 g/mol HCl; 36.45 g/mol What is the limiting reagent in this scenario?
In one experiment, 12.0 g of PCl5 was slowly added to 15.0 g of H2O according to the following balanced chemical equation:
Pcl5(s)+4H2O(l) ---> H3PO4 (aq) + 5HCl (g)
The molar masses for each compounds in the equation are as follows;
PCl5: 208.224 g/mol
H2O: 18.015 g/mol
H3PO4: 97.994 g/mol
HCl; 36.45 g/mol
What is the limiting reagent in this scenario?
Note! Make an accurate claim: The limiting reagent in this scenario is _____. Provide additional details and use subject specific language.
Cite evidence from what's given to you in the problem that supports your answer, hint: look at that equation!. Lastly! Fully connect the evidence to the claim/your answer to the question. Include subject specific language in your reasoning.
Your written response should LOOK exactly like the response attached. The other image is there to show how you would figure out the limiting reagent.

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