We fused 10 g of copper and 9 g of sulfur. determine the mass and quantity of the obtained sulfide.

Let’s write down given:

m (Cu) = 10 g

m (S) = 9 g

m (CuS) -?

Solution:

1) Let’s compose the equation for the reaction of interaction of copper with sulfur:

Cu + S = CuS

2) Find the amount of substance copper and sulfur:

M (Cu) = 64 g / mol

ν = m / M

ν (Cu) = 10 g / 64 g / mol = 0.16 mol

M (S) = 32 g / mol

ν (S) = 9 g / 32 g / mol = 0.28 mol

In the reaction, 1 mol of copper reacts with 1 mol of sulfur, therefore, 0.16 mol of sulfur will react with 0.16 mol of copper. Sulfur is in excess, which means that we calculate the mass of sulfur sulfide for copper:

M (Cu S) = 64 + 32 = 96 g / mol

From 64 g (Cu) 96 g (Cu S) is formed

From 10 g (Cu) x g (Cu S) is formed

X = 10g * 96g / 64g

X = 15 g (Cu S)

Answer: 15 g (Cu S)