Answer to Question #121943 in General Chemistry for achol bol

Question

. During the formation of titanium (III) oxide, 40.0 g of titanium reacts with 20.0 g of oxygen.
What mass of compound should be produced?

Answer 1

Balanced chemical equation is

2Ti + 3/2O₂ =>Ti₂O₃

given Ti is 40g and O₂ is 20 grams

no. of moles of Ti = 40/48 = 0.833moles

no. of moles of O₂ = 20/32= 0.625 moles

now according to stiochiometry of reaction

ratio of moles of reactants must be 2:(3/2)

but according to given amount ratio is 0.833:0.625 which is equivalent to 2:(3/2)

so there is no limiting reagent

since 2 moles of Ti react with 1.5 moles of O₂ to produce 1 mole of product

so for 0.833 moles of Ti react with 0.625 moles of O₂ to produce (0.833/2) moles of product

mass of product obtained = no. of moles* M.W

where M.W is molecular weight of Ti₂O₃ (=144gmol^-1)

mass of product = (0.833/2)*144 grams

= 59.97 grams