When 320.62 grams of Bi2O3 react with excess of C, how many moles of Bi form?
I know I need to use stoichiometry, but I can't quite seem to get the correct balanced equation.
2Bi2O3 + 3C ==> 3CO2 + 4Bi
Thank you
To solve this problem, we need to start by balancing the equation. Let's determine the balanced chemical equation for the reaction between Bi2O3 and C.
The reactants are Bi2O3 and C, and the products are Bi and CO2. The unbalanced equation is:
Bi2O3 + C → Bi + CO2
To balance the equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation. Let's balance each element one by one.
Balance oxygen (O) first. The Bi2O3 has three oxygen atoms while CO2 has two oxygen atoms, so we place a coefficient of 3 in front of CO2:
Bi2O3 + C → Bi + 3CO2
Now, let's balance the bismuth (Bi) and carbon (C) atoms. Since there are two bismuth atoms on the left, we need to place a coefficient of 2 in front of Bi:
2Bi2O3 + C → 4Bi + 3CO2
The balanced equation is:
2Bi2O3 + C → 4Bi + 3CO2
Now that we have the balanced equation, we can use stoichiometry to determine the number of moles of Bi formed.
According to the balanced equation, the mole ratio of Bi2O3 to Bi is 2:4, which simplifies to 1:2. This means that for every 2 moles of Bi2O3, we will get 4 moles of Bi.
To determine the number of moles of Bi formed, we need to convert the mass of Bi2O3 to moles. The molar mass of Bi2O3 is calculated by adding the atomic masses of bismuth (Bi) and oxygen (O):
Bi: 208.98 g/mol
O: 16.00 g/mol
Molar mass of Bi2O3 = 2 * (Bi) + 3 * (O) = 2 * 208.98 g/mol + 3 * 16.00 g/mol = 465.96 g/mol.
To convert grams of Bi2O3 to moles, we divide the given mass (320.62 g) by the molar mass of Bi2O3:
moles of Bi2O3 = mass of Bi2O3 / molar mass of Bi2O3 = 320.62 g / 465.96 g/mol.
Now, we can use the mole ratio to find the number of moles of Bi:
moles of Bi = moles of Bi2O3 * (4 moles of Bi / 2 moles of Bi2O3).
Simply substitute the value of moles of Bi2O3 into the equation:
moles of Bi = (320.62 g / 465.96 g/mol) * (4 moles of Bi / 2 moles of Bi2O3).
By calculating this expression, you will find the number of moles of Bi formed in the reaction.