Question

The reaction of Cr203 with silicon metal at high temperatures will make chromium metal. 2Cr,03 (s) +3Si(s) → 4Cr(1) +3SiO2 (s) The reaction is begun with 98.00 g of Si and 135.00 g of Cr203

Answer 1

1)

Molar mass of Cr2O3 = 2*MM(Cr) + 3*MM(O)

= 2*52.0 + 3*16.0

= 152 g/mol

mass of Cr2O3 = 135.0 g

we have below equation to be used:

number of mol of Cr2O3,

n = mass of Cr2O3/molar mass of Cr2O3

=(135.0 g)/(152 g/mol)

= 0.8882 mol

Molar mass of Si = 28.09 g/mol

mass of Si = 98.0 g

we have below equation to be used:

number of mol of Si,

n = mass of Si/molar mass of Si

=(98.0 g)/(28.09 g/mol)

= 3.489 mol

we have the Balanced chemical equation as:

2 Cr2O3 + 3 Si ---> 4 Cr + 3 SiO2

2 mol of Cr2O3 reacts with 3 mol of Si

for 0.8882 mol of Cr2O3, 1.332 mol of Si is required

But we have 3.489 mol of Si

so, Cr2O3 is limiting reagent

Answer: Cr2O3

2)

we will use Cr2O3 in further calculation

From balanced chemical reaction, we see that

when 2 mol of Cr2O3 reacts, 3 mol of Si reacts

mol of Si reacted = (3/2)* moles of Cr2O3

= (3/2)*0.8882

= 1.332 mol

mol of Si remaining = mol initially present - mol reacted

mol of Si remaining = 3.489 - 1.332

mol of Si remaining = 2.157 mol

Molar mass of Si = 28.09 g/mol

we have below equation to be used:

mass of Si,

m = number of mol * molar mass

= 2.157 mol * 28.09 g/mol

= 60.6 g

Answer: 60.6 g

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