what mass of grams of hydrogen sulfide will be required to participate 15 g of copper sulphide from a copper (ii) traoxosulphate
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Answer:
103.9 g
Explanation:
- C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
First <u>we convert 54.0 g of propane (C₃H₈) into moles</u>, using its <em>molar mass</em>:
- 54.0 g ÷ 44 g/mol = 1.23 mol C₃H₈
Then we <u>convert 1.23 moles of C₃H₈ into moles of CO₂</u>, using the <em>stoichiometric coefficients</em>:
- 1.23 mol C₃H₈ *
= 3.69 mol CO₂
We <u>convert 3.69 moles of CO₂ into grams</u>, using its <em>molar mass</em>:
- 3.69 mol CO₂ * 44 g/mol = 162.36 g
And <u>apply the given yield</u>:
- 162.36 g * 64.0/100 = 103.9 g