Question

What is the theoretical yield of bismuth when 198 g of Bi2O3 reacts with excess carbon as shown below? Bi2O3 C(s)>Bi(s) + CO(g) (not balanced)

Answer 1

The theoretical yield of bismuth in the given reaction is 0.848 moles or 177.2 grams.

Step-by-step explanation:

To determine the theoretical yield of bismuth (Bi) in the reaction between 198 g of Bi2O3 and excess carbon, we need to balance the equation first. The balanced equation for the reaction is:

Bi2O3 + 3C → 2Bi + 3CO

From the balanced equation, we can see that for every mole of Bi2O3, we get 2 moles of Bi. To calculate the theoretical yield, we need to convert the given mass of Bi2O3 to moles using its molar mass (465.96 g/mol) and then use the stoichiometry to find the moles of Bi.

Mass of Bi2O3 = 198 g

Molar mass of Bi2O3 = 465.96 g/mol

Moles of Bi2O3 = (198 g) / (465.96 g/mol) = 0.424 mol

Moles of Bi = 2 * (0.424 mol) = 0.848 mol

The theoretical yield of bismuth in this reaction is 0.848 moles or you can convert it to grams using the molar mass of bismuth (208.98 g/mol) to get the theoretical yield in grams.

Answer 2

179.4306 g

Step-by-step explanation:

Given that:

Mass of
`Bi_2O_3` = 198 g

Molar mass of
`Bi_2O_3` = 465.96 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`Moles= (198\ g)/(465.96\ g/mol)`

`Moles\ of\ Bi_2O_3= 0.4293\ mol`

From the balanced reaction,

`Bi_2O_3+3C\rightarrow 2Bi+3CO`

1 mole of
`Bi_2O_3` on reaction produces 2 moles of bismuth

So,

0.4293 mole of
`Bi_2O_3` on reaction produces 2 × 0.4293 moles of bismuth

Moles of bismuth = 0.8586 moles

Molar mass of bismuth = 208.9804 g/mol

So, mass of bismuth = Moles × Molar mass = 0.8586 × 208.9804 g = 179.4306 g