Given your answers from Part A and B, determine the limiting reactant if 10 0 g of Cr and 10.0 g of O₂ reacted in the following reaction 4Cr(s)+30₂(g) →2Cr₂O3(s) Y 0₂ Cr₂O₂ Cr Subm Previous Answers ✓ Correct The amount of Cr₂O₂ that can be formed from 10.0 g of Cr must be compared to the amount of Cr₂Oy that can be formed from 10.0 g of O₂ Whichever reactant forms the least amount of Cr₂O₂ is the limiting reactant, as that reactant was completely consumed. From Parts A and B 146 g of Cr₂O₂ is produced by 10 0 g of Cr and 31.7 g of Cr₂O₂ is produced by 100 g of O₂ 4 Therefore, since the reactant Cr produced the least amount of product Cr₂O₂, the limiting reactant is 10.0 g of Cr Part D What is the actual number of grams of Cr₂O3 that would be produced in Part C? Express your answer with one decimal place and with appropriate units.

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Mole Concept

Mole concept is a method to determine the amount of any substance that consists of some elemental particles using a grouping unit called ‘mole’. The mole concept enables one to handle the large amount of small elementary particles. The mole concept can be used in many calculations related to mole calculation, molar amount, mole ratio, and so on, which are significant in chemistry.