For the reaction, calculate how many grams of the product form when 17.4 g of Sr completely...

Question:

For the reaction, calculate how many grams of the product form when 17.4 g of Sr completely reacts. Assume that there is more than enough of the other reactant. 2 Sr(s) + O_2(g) ---> 2 SrO(s) mSrO =

Limiting Reactant

This is the reactant fully used up in the reaction. In this case, it states that there is more than enough which means that it is in excess, in this case, the oxygen gas

Answer and Explanation: 1

Compute for mass of the product

{eq}mass\: SrO=(17.4g\: Sr)(\frac{1mole}{87.62g})(103.62\frac{g}{mol})=20.58g {/eq}


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Limiting Reactants & Calculating Excess Reactants

from

Chapter 9 / Lesson 5

Learn how to find the limiting and excess reactants in a chemical reaction. See example problems that calculate the limiting and excess reactants.


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