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The reaction of Cr203 with silicon metal at high temperatures will make chromium metal. 2Cr,03 (s)...

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

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
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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

Feel free to comment below if you have any doubts or if this answer do not work. I will edit it if you let me know

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