Question

Example:. 4. Use: Cu Copper + S Sulfuro ' Cus Copper(II)sulfide a) If we start with 20 gram of copper and 20 gram of sulfur, how many grams of Copper (II) sulfide is produced? b) If we start with 100 gram of copper, how much sulfur in grams is required and how many grams of Copper (II) sulfide is produced? Example: Calculate percentage yield of Magnesium oxide when 200.0 grams of oxygen gas reacts with 70.0 grams of Magnesium metal and produces 100.6 grams of Magnesium oxide.

Answer 1

Solution 4 7 Civen- Cut S – copper sulfus Cus copper (II) sulfide (a) Civen - mass of copper = 20 gram mass of sulfus = 20 gra Molecular mass of molecular mass of So, Molecular mass of copper = 63.55 g / mal safer = 32 g 1 mol copper (1) sulfide= (63.55+ 32) = 95.55 g/mol We have moles of to find out copper = the colh mole component's mole. given mass of an an Molecular mass of 20g 6 3.55 gemoe = 0.3147 mole hou present or available

no. of moles of sulfur = given mass of S. molecular mass of s = 20g = 0.625 male 329/mol ts= 0.625 male So l'han cr I moles of copper present is ess than suffus present so, copper is limiting agent. Here 1 male of copper produce - I male of cus. So, 0.3147 moll of co produce = 0.3147 male. of cus or 63.66 g of cu produce = 95.55 g pf cus- So, Ig of ce produce = 95.55 63.55 so g of cus 20g of cu produce = 95.55x20 g 63.55 U – 30.071 g of cas Ansso, 30.1g of copper (17) sulfide is produced.

6 Cooper 100 gram. 1 (6) = Civen - moss of copper = loog Here 63.55 gram of cu react with = 32 grom of so, ag of an react with 232 g of s 100g of an will react with =( 32 x 100) = 50.35 g ofs sulfur is required = 50.35 gram. Ans 63.55 g of cu produce = 95.559 of a swepide I g of a produce = 95.55 a 100 g of a produce = 95.55x100 g 63.55 cus 63.55 = 150.35 g Of cus is produced = 150.35 gram Ans copper (II) sulfide