need help plz Or32332) HWsz 5.63.3 6. 212 021 5.0 points Consider the following

Question

need help plz

Or32332) HWsz 5.63.3 6. 212 021 5.0 points Consider the following reaction How mudi CaCN2 is noded to ptodu, 2.6 moles of N113 if the percent yield is 65.8%? 1.9.57 2. 6.30 3. 4.14 4. 332 5.0.315 8 6. 767 g 7, 0.479 8. 501 022 5.0 points Consider the reaction: The reaction has a 291% yield How many grams of Al2S3 are needed to react with 30.0 moles of IICI? The molar m AlS in 150,16 ams/mold The molar mass of AlCla 34.080 grams/mole 1, 467 grams 2. 1.20 grams 3. 2080 gramn 4. 180. gras grams/moleThe molar masn of HCI is 36.4 133.33 grams/mole) The molar mass of Hz

Explanation / Answer

21) from equation, 1 mol CaCN2 = 2 mol NH3

   No of mol of NH3 produced = 12.6 mol

percent yield = practical yield / theoretical yield*100

   65.8 = (12.6/x)*100

x = theoretical yield = 19.15 mol

so that,

Amount of CaCN2 required = (19.15/2)*80.102

    = 767 g

answer: 6) 767 g

22) from the equation, 1 mol Al2S3 = 6 mol HCl

No of mol of Al2S3 required react completely with 30 mol HCl

            = 30/6 = 5 mol   ( if 100% yield)

percent yield = practical yield / theoretical yield*100

       29.1 = 5/x*100

    x = actual No of mol of Al2S3 required = 17.18 mol

Amount of Al2S3 = 17.18*150.16 = 2580 grams

answer: 3) 2580 g

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