Answer:
correct option: 1
excess reactant sulpher and mass left=10g
Explanation:
first write ha balance chemical equation:
[tex]2 S + 3 O_2 \rightarrow 2 SO_3[/tex]
[tex]mole=\frac{given \, mass}{molar\,mass}[/tex]
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
