Solid bismuth oxide can react with carbon to form bismuth metal and carbon monoxide. How many grams of bismuth oxide reacted if 60.7 grams of bismuth is formed?

Question

Naomi Aronson · Accepted Answer

The answer is #67.6# #g#. First, utilize the chemical equation that is balanced. #Bi_2O_(3(s)) + 3C_((s)) -> 2Bi_((s)) + 3CO_((g))# We know the molar masses of #Bi# (#209g/(mol)#) and #Bi_2O_3# (#466g/(mol)#), and the mole-to-mole ratio between #Bi# and #Bi_2O_3#, so we could say that #60.7gBi* (1 mol e Bi)/(209g) * (1 mol e Bi_2O_3)/(2 mol es Bi) * (466g)/(1 mol e Bi_2O_3) = 67.6g Bi_2O_3#

Cameron Andrews · Answer

undefined To find the grams of bismuth oxide reacted, we need to use stoichiometry based on the balanced chemical equation for the reaction. The balanced equation for the reaction is:

Bi2O3 + 3C -> 2Bi + 3CO

First, we need to find the molar mass of Bi2O3, which is 2*208.98 (molar mass of Bi) + 3*16.00 (molar mass of O) = 465.96 g/mol.

Using stoichiometry, we can calculate the moles of Bi produced from the given mass:

Moles of Bi = Mass of Bi / Molar mass of Bi = 60.7 g / 208.98 g/mol = 0.290 moles of Bi

From the balanced equation, we can see that the mole ratio between Bi2O3 and Bi is 1:2. So, the moles of Bi2O3 reacted would be half of the moles of Bi produced:

Moles of Bi2O3 = Moles of Bi / 2 = 0.290 moles / 2 = 0.145 moles of Bi2O3

Finally, we can find the grams of Bi2O3 reacted using its molar mass:

Mass of Bi2O3 = Moles of Bi2O3 * Molar mass of Bi2O3 = 0.145 moles * 465.96 g/mol = 67.58 grams

Therefore, approximately 67.58 grams of bismuth oxide reacted.