When copper is heated with an excess of sulfur, copper(l) sulfide is formed. In a given experiment, 0.0970 moles of copper was heated
with excess sulfur to yield 5.59 g copper(1) sulfide. What is the
percent yield?

72.4%

Explanation:

Step 1: Write the balanced equation

2 Cu + S ⇒ Cu₂S

Step 2: Calculate the theoretical yield, in moles, from 0.0970 moles of Cu

The molar ratio of Cu to Cu₂S is 2:1.

0.0970 mol Cu × 1 mol Cu₂S/2 mol Cu = 0.0485 mol Cu₂S

Step 3, Convert the theoretical yield to mass

The molar mass of Cu₂S is 159.16 g/mol.

0.0485 mol × 159.16 g/mol = 7.72 g

Step 4: Calculate the percent yield

We will use the following expression.

%yield = experimental yield/theoretical yield × 100%

%yield = 5.59 g/7.72 g × 100% = 72.4%

it would be a gas formation

it is a solid

explanation: