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

Answer:

179.4306 g

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:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{198\ g}{465.96\ g/mol}$

$Moles\ of\ Bi_2O_3= 0.4293\ mol$

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