Question

60.0 g O2 and 50.0 g S are reacted according to the equation 2 S + 3 O2 → 2 SO3 . Which reactant is in excess and by how many grams? 1. S; 10.0 g 2. O2; 10.0 g 3. S; 20.0 g 4. S; 24.8 g 5. O2; 20.0 g 6. O2; 24.8 g

Answer 1

Answer:

correct option: 1

excess reactant sulpher and mass left=10g

Explanation:

first write ha balance chemical equation:

$2 S + 3 O_2 \rightarrow 2 SO_3$

$mole=\frac{given \, mass}{molar\,mass}$

Given mass of sulfur=50g

molar mass=32 g/mol

mole of sulfur=1.5625 mol

given mass of O_2=60g

molar mass=32 g/mol

mole of oxygen=1.875mol

from the above balanced equation it is clearly that;

2 mole of sulfur reacts completely with 3 mole of Oxygen

1 mole of sulfur reacts completely with 1.5 mole of Oxygen

1.5625 mole of sulfur will react with 1.5×1.5625 (2.34375 mol)mole of oxygen

but we have only 1.875 mole of oxgygen hence sulpfur will be the excess reactant and oxygen will be thelimiting reactant

therefore,

3 mole of Oxygen reacts completely with 2 mole of sulfur

1 mole of Oxygen reacts completely with 2/3 mole of sulfur

so

1.875 mole of Oxygen will react completely with (2/3)×1.875(1.25mol) mole of sulfur

mole of sulfur used =1.25 mol

left mole of sulfur=1.5625-1.25=0.3125

mass left=0.3125×32g=10g

excess reactant sulpher and mass left=10g