Question

21 Question (1 point) Chromium(III) oxide reacts with silicon as indicated by the balanced equation below. 2Cr,0, (s) +35i(s) — 4Cr(s) +3Si0,(s) When 95.00 g Cr203 is reacted with 119.0 g Si, it is determined that Cr2O3 is the limiting reagent. 1st attempt . See Periodic Table See Hint If the reaction begins with 95.00 grams of Cr2O3 and 119.0 grams of silicon, how many grams of the excess reactant will be used? g of Si used

Answer 1

Consider reaction , 2 Cr₂O₃ (s) + 3 Si (s) $\rightarrow$ 4 Cr (s) + 3 SiO ₂ (s)

From above reaction, we can write 2 mol  Cr₂O₃ $\equiv$ 3 mol Si $\equiv$ 4 mol Cr $\equiv$ 3 mol SiO ₂

Molar mass of Si = 28.08 g/mol

Molar mass of Cr₂O₃ = ( 2 $\times$ 52.00 ) + ( 3 $\times$ 16.00 ) = 152 g/mol

$\therefore$ 2 $\times$ 152 g Cr₂O₃$\equiv$3 $\times$ 28.08 g Si

95.00 g Cr₂O₃ $\equiv$3 $\times$ 28.08 $\times$ 95.00 / (  2 $\times$ 152) g Si

95.00 g Cr₂O₃ $\equiv$ 26.325 g Si

i e 95.00 g  Cr₂O₃ reacts with 26.325 g Si .

ANSWER : If the reaction begins with 95.00 g  Cr₂O₃ and 119.0 g Si , how many grams of excess reagent used ?

ANSWER : 26.33 g