can someone please help me with this asap?

Question

. Copper reacts with sulfur to form copper (I) sulfide according to the following reaction:

  

2 Cu + S  →  Cu2S

When 98.6 g of copper reacts with 41.2 grams of sulfur 

1. What is the % yield if 96.2 grams of product are formed?
2. Determine grams of excess reactant left over at the end of the reaction.

Answer 1

There are two conversion factors to consider.

Mol mass Cu2S / 2 x mol mass Cu = 159.16 g Cu2S / 2 x 63.55 g Cu = 1.252

and mol mass Cu2S / mol mass S = 159.16 g Cu2S / 32.06 g S = 4.964

Now multiply 1.252 x g Cu given = 1.252 x 98.6 = 123.4 g Cu2S predicted.

Now multiply 4.964 x 41.2 g S given = 204.5 g Cu2S predicted.

As fewer grams of Cu2S are predicted with the first conversion factor, the limiting reactant is Cu.

The %yield is mass CuS obtained / mass Cu predicted x 100 = 96.2 / 123.4 x 100 = 78.0%

According to the balanced equation, the mass ratio of Cu to Cu2S is 2x 63.55 / 159.16 = 0.799

If only 96.2 g of Cu2S was produced, the mass of Cu actually reacted is 0.799 x 96.2 = 76.9 g Cu.

If you began with 98.6 g Cu, and only 76.9 g were used, there are 98.6 - 76.9 = 21.7 g left over.

As 96.2 g Cu2S were produced, the mass of S actually reacted is mol mass S / mol mass Cu2S x 96.2 g Cu2S = 32.06 / 159.16 x 96.2 = 19.4 g S.

As you began with 41.2 g S, there are 41.2 - 19.4 = 21.8 g S left over.

Please check all math.