Question

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 *

Answer 1

Limiting reactant is CS2

% yield = 131.8 %

Step-by-step 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 ---
`(35)/(76.14)` moles

= 0.4597 moles

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

= 27.6 g of C.

Percentage yield (%) =
`(actual- yield)/(theoretical- yield)`

=
`(36.4)/(27.6) * 100` =131.9%