Asked Dec 11, 2019

If 65.0 g of Al2S3 reacts with excess H2O at 90.0°C and 0.987 atm, what volume of H2S gas will be produced?


Expert Answer

Step 1

Given information:


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Mass of Al,S, = 65g Molar massof AI,S, =150.158g / mol 65g so, mole of Al,S, 150.158g / mol =0.4328 moles

Step 2

Moles of H2S formed can be calculated as


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Al,S,(s) +6H,O(1) →3H,S(g)+2Al(OH),(s) 3 moles H,S = 0.4328moles Al,S, x 1mol Al S, n=1.2984moles. Temperature=90 +273=363K = 0.987 atm Pressure


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