How Much Chromium(III) Oxide Is Needed for 439g of Chromium Sulfide?

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Chromium(III) Oxide (Cr2O3) reacts with Hydrogen Sulfide (H2S) to produce Chromium Sulfide (Cr2S3) and water, following the reaction equation Cr2O3(s) + H2S(g) = Cr2S3(s) + H2O(l). To produce 439g of Cr2S3, the required moles of Cr2O3 can be determined using a 1:1 mole ratio between Cr2O3 and Cr2S3. The calculation involves first finding the moles of Cr2S3 in 439g, which directly indicates the moles of Cr2O3 needed. Subsequently, the mass of Cr2O3 can be calculated from the moles obtained in the first part. Accurate calculations will yield the necessary amounts for both parts of the inquiry.
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1. Chromium(III) Oxide reacts with Hydrogen Sulfide(H2S) gas to form Chromium Sulfide and water.

CR2O3(s) + H2S(g) = Cr2S3(s) + H2O (l)

to produce 439g of Cr2S3.

a. How many moles of Cr2O3 are required?

b. How many grams of CR2O3 are required?



2. Not sure how to set this up



3. All I could think of was finding the total grams of CR2O3 and dividing it from 439g's to find part a, but I think I was wrong
 
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Identify the ratios from the written reaction. Calculate the formula weights for the relevant compounds.
 
For part A: Find out how moles of Cr2S3 are in 439 g and the answer is how many moles of Cr2O3 because the mole:mole ratio is 1:1.

For part B: Calculate how many grams of Cr2O3 are in your answer to part a
 

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