Bismuth oxide reacts with carbon to form bismuth metal: bi2o3(s) + 3c(s) → 2bi(s) + 3co(g) when 689 g of bi2o3 reacts with exces
s carbon, (a) how many moles of bi form? 2.957 mol bi (b) how many grams of co form? g co
2 answers:
CO mass = 4,435. 18 = 79,839 grams
Stokiometry in Chemistry learns about chemical reactions mainly emphasizing quantitative, such as calculation of volume, mass, number, which is related to the number of ions, molecules, elements etc.
In chemical calculations, the reaction can be determined, the number of substances that can be expressed in units of mass, volume, mole, or determine a chemical formula, for example the substance level or molecular formula of hydrate.
In stockiometry therein includes
relative atomic mass (Ar) and relative molecular mass (Mr)
Mr. AxBy = (x.Ar A + y. Ar B)
Reactions that occur:
Bi₂O₃ (s) + 3C (s) → 2Bi (s) + 3CO (g)
- We specify mole Bi₂O₃
Mr Bi₂O₃ = 2. ar bi + 3. Ar O
Mr Bi₂O₃ = 2. 209 + 3. 16
Mr. Bi₂O₃= 466
mole Bi₂O₃ = gram / Mr
mole = 689/466
mole 1.4785
- A. Comparison of Bi reaction coefficients: Bi₂O₃ = 1: 2, then Bi moles = 2. 1,4785 = 2, 957
Comparison of Bi reaction coefficients: Bi₂O₃ = 1: 2,
- B. While the number of moles CO = 3 x 1.4785 = 4,435
Mr. CO = 12 + 16 = 18
mass CO = mole. Mr
CO mass = 4,435. 18 = 79,839 grams
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Answer:
10 molecules of NH₃.
Explanation:
N₂ + 3H₂ --> 2NH₃
As the N₂ supply is unlimited, what we need to do to solve this problem is <u>convert molecules of H₂ into molecules of NH₃</u>. To do so we use the <em>stoichiometric coefficients</em> of the balanced reaction:
- 15 molecules H₂ *
= 10 molecules NH₃
10 NH₃ molecules could be prepared from 15 molecules of H₂ and unlimited N₂.