What is the percent yield of C when 35 g CS₂ are reacted and produce 36.4 g C. * LABEL THE LIMITING REACTANT IN YOUR WORK.* Use the following balanced chemical equation to solve the problem: CS₂ + 4 CO --> 5 C + 2 SO₂ *

Respuesta :

Answer:

Limiting reactant is CS2

% yield = 131.8 %

Explanation:

Using the alanced equation given, Mole ratio of C to CS2 is 1 : 5.

That is, for every 1 mole of CS2 used, 5 moles of Carbon is produced.

35g of CS2 --- [tex]\frac{35}{76.14}[/tex] moles

= 0.4597 moles

⇒ mass of C produced = 5 × 0.4597 × 12 g/mol

= 27.6 g of C.

Percentage yield (%) = [tex]\frac{actual- yield}{theoretical- yield}[/tex]

= [tex]\frac{36.4}{27.6} * 100[/tex] =131.9%