Answer to Question #306081 in General Chemistry for Mark Herrera

Question

For the following reaction, 10.6 grams of sulfur are allowed to react with 22.8 grams of carbon monoxide .

sulfur ( s ) + carbon monoxide ( g )  sulfur dioxide ( g ) + carbon ( s )

What is the maximum amount of sulfur dioxide that can be formed?  

What is the FORMULA for the limiting reagent?

What amount of the excess reagent remains after the reaction is complete?

Answer 1

S + 2CO = SO₂ + 2C

First, look for the amount of substance of sulfur:

n(S) = m / M

n(S) = 14.8 g/32 g / mol = 0.4625 mol

n(CO) = m (CO) / M (CO)

M(CO) = 12 + 16 = 28 g/mol

n(CO) = 19.9 g/28 g/mol = 0.71 mol

S in excess, so for calculating we take CO:

n(SO₂) = n(CO)/2 = 0.71 mol/2 = 0.355 mol

m(SO₂) = M(SO₂)*n(SO₂)

M(SO₂) = 32 + 16*2 = 64 g/mol

m(SO₂) = 64 g/mol * 0.355 mol = 22.74 g