Question

The reaction of Cr2O3 with silicon metal at high temperatures will make chromium metal. 2CrO3(s) + 3Si(s)----> 4Cr(l) + 3SiO2 (s) The reaction is begun with 92.00 g of Si and 112.00 g of Cr2O3. How many grams of the excess reactant are left after the reaction is complete?

Answer 1

Answer: 51.45 grams of excess reagent is left after the completion of reaction.

Explanation: For the calculation of moles, we use the formula:

`Moles=\frac{\text{Given mass}}{\text{Molar mass}}` ....(1)

• For Si

Given mass = 92 grams

Molar mass = 28g/mol

Putting values in equation 1, we get:

`Moles=(92g)/(28g/mol)=3.285moles`

• For
  `Cr_2O_3`

Given mass = 112 grams

Molar mass = 116g/mol

Putting values in equation 1, we get:

`Moles=(112g)/(116g/mol)=0.965moles`

The reaction follows:

`2Cr_2O_3(s)+3Si(s)\rightarrow 4Cr(l)+3SiO_2(s)`

By Stoichiometry,

2 moles of
`Cr_2O_3` reacts with 3 moles of silicon

So, 0.965 moles of
`Cr_2O_3` reacts with =
`(3)/(2)* 0.965` = 1.4475 moles of Silicon.

As, the moles of silicon is more than the required amount and is present in excess.

So, the excess reagent for the reaction is Silicon.

Moles of silicon remained after reaction = 3.285 - 1.4475 = 1.8375 moles

To calculate the amount of Silicon left in excess is calculated by using equation 1:

`1.8375=\frac{\text{Given mass}}{28}`

Amount of Silicon in excess will be 51.45 grams.