Answer: 2.25 moles of C, 40.6 grams of Sb, 77%, and
How many moles of C are needed to produce 4.5 moles of CO?
Answer: 2.25 moles of C
Step-by-step explanation: To answer this question, we need to use a mole ratio, which is a stoichiometric relationship between the amounts in moles of any two substances in a chemical reaction. The balanced equation for the reaction is:
\ce {C} (s) + \ce {O2} (g) \rightarrow \ce {CO} (g) + \ce {CO2} (g) C(s) + OX 2(g) → CO(g) + COX 2(g)
The mole ratio between C and CO is 1:1, which means that for every mole of C that reacts, one mole of CO is produced. Therefore, to produce 4.5 moles of CO, we need 4.5 moles of C.
How many grams of antimony (Sb) are produced when 56.7 grams of antimony oxide (Sb2O3) are used up completely?
Answer: 40.6 grams of Sb
Step-by-step explanation: To answer this question, we need to use composition stoichiometry, which is the calculation of quantities by weight in a reaction described by a balanced equation. The balanced equation for the reaction is:
\ce {Sb2O3} (s) + \ce {C} (s) \rightarrow \ce {Sb} (s) + \ce {CO} (g) SbX 2OX 3(s) + C(s) → Sb(s) + CO(g)
To find the mass of Sb produced from a given mass of SbX 2OX 3, we need to convert the mass of SbX 2OX 3 to moles using its molar mass, then use the mole ratio between SbX 2OX 3 and Sb to find the moles of Sb produced, and then convert the moles of Sb to mass using its molar mass. The molar masses of SbX 2OX 3 and Sb are 291.5 g/mol and 121.8 g/mol, respectively. The mole ratio between SbX 2OX 3 and Sb is 1:2, which means that for every mole of SbX 2OX 3 that reacts, two moles of Sb are produced.
A chemist is able to collect 18.3 grams of Sb from the reaction of 42 grams of C with excess Sb2O3. What is the chemist’s percent yield of Sb?
Answer: 77%
Step-by-step explanation: To answer this question, we need to use the concept of percent yield, which is the ratio of the actual yield (the amount of product obtained from a reaction) to the theoretical yield (the maximum amount of product that can be obtained from a reaction) expressed as a percentage. The balanced equation for the reaction is:
\ce {Sb2O3} (s) + \ce {C} (s) \rightarrow \ce {Sb} (s) + \ce {CO} (g) SbX 2OX 3(s) + C(s) → Sb(s) + CO(g)
A chemist has 62 grams of Sb2O3 and 33 grams of C available to produce Sb. What is the limiting reagent?
Answer: C is the limiting reagent
Step-by-step explanation: To answer this question, we need to use the concept of limiting reagent, which is the reactant that is completely used up in a reaction and thus determines when the reaction stops. The balanced equation for the reaction is:
\ce {Sb2O3} (s) + \ce {C} (s) \rightarrow \ce {Sb} (s) + \ce {CO} (g) SbX 2OX 3(s) + C(s) → Sb(s) + CO(g)
To find the limiting reagent, we need to compare the amounts of SbX 2OX 3 and C in moles and use the mole ratio between them from the balanced equation. The mole ratio between SbX 2OX 3 and C is 1:1, which means that for every mole of SbX 2OX 3 that reacts, one mole of C is needed. Therefore, if we have more moles of one reactant than the other, that reactant is in excess and the other one is limiting. To find the moles of SbX 2OX 3 and C, we need to use their molar masses. The molar masses of SbX 2OX 3 and C are 291.5 g/mol and 12.0 g/mol, respectively.
Since we have more moles of C than SbX 2OX 3, C is in excess and SbX 2OX 3 is limiting. However, another way to find the limiting reagent is to use a formula based on the mole ratio:
Limiting Reagent Formula
- Determine the amount of each reactant in moles.
- In the balanced chemical equation, divide the actual number of moles of each reactant by its stoichiometric coefficient.
- The limiting reactant is the one with the lowest mole ratio.
- Using this formula, we can find the limiting reagent.
Since 0.213 < 2.75, SbX 2OX 3 is the limiting reagent.
Hope this helps, and have a great day! =)