Elemental antiomony made by these reactions down below:
Sb2S3 + 5 O2 - > Sb2O4 + 3 SO2
Sb2O4 + 4 C - > 2 Sb + 4 CO
how many grams O2 needed to prepare 1.00g of Sb?
how many grams Sb formed when 84.9g of Sb2S3 and 39.0g of O2 r used in first reaction and product formed is used in second reaction?
i not understand these.
figure how many moles of Sb is one gram.
Then, you need 1/2 that number of moles of Sb2O4. That is the number of moles of Sb2O4 in the first reaction. Then you need 5 times that number of moles of O2.
putting it together, you figure the moles of Sb. Then you need 5/2 of that of oxygen for the first equation.
Now on the second q, it is totally different. You will have to figure the limiting reactant in the equation in the reaction. Figure the moles of Sb2S3, and O2 from the masses given. Then, do you have 5 times as much O2? If you dont, then oxygen is limiting, and you figure the reaction based on just the moles of oxgen. If you have more than 5x , then Sb2S2 is limiting, and you figure the reaction based on that amount.
i actually doing this first time and i real not get how to do this how i find out how many mole Sb in one gram?
moles = grams/molar mass
To answer the given questions, let's break down each step and calculate the quantities of substances involved.
Question 1: How many grams of O2 are needed to prepare 1.00g of Sb?
In the first reaction, the balanced equation shows that for every 5 moles of O2 used, we obtain 3 moles of SO2. Similarly, the second reaction tells us that for every 4 moles of C used, we obtain 2 moles of Sb. We need to establish a bridge between these two reactions.
1. Determine the molar mass of Sb.
By looking up the molar mass of Sb on the periodic table, we find that it is approximately 121.7 g/mol.
2. Use stoichiometry to find the moles of Sb.
Given that we have 1.00g of Sb, we can calculate the number of moles using the formula:
moles = mass (grams) / molar mass (g/mol)
moles = 1.00g / 121.7 g/mol
moles ≈ 0.00821 mol
3. Use the stoichiometry of the reaction equations to find the moles of O2 required.
From the balanced equation, we see that 2 moles of Sb are produced for every 5 moles of O2 used. Therefore, we can set up the following ratio:
5 moles O2 / 2 moles Sb = x moles O2 / 0.00821 mol
Solving for x, we get x ≈ 0.0205 mol.
4. Convert the moles of O2 to grams.
Now that we know the moles required, we can calculate the gram amount using the molar mass of O2, which is approximately 32.0 g/mol.
grams = moles × molar mass
grams = 0.0205 mol × 32.0 g/mol
grams ≈ 0.656 g
Therefore, approximately 0.656 grams of O2 are required to prepare 1.00 gram of Sb.
Question 2: How many grams of Sb are formed when 84.9g of Sb2S3 and 39.0g of O2 are used in the first reaction, and the product is used in the second reaction?
We will follow a similar process to answer this question.
1. Determine the limiting reactant.
To identify the limiting reactant, we need to compare the moles of Sb2S3 and O2 used. Let's calculate the moles of each substance using their molar masses.
For Sb2S3:
moles = mass (grams) / molar mass (g/mol)
moles = 84.9g / (121.7 g/mol + 2 × 32.1 g/mol)
moles ≈ 0.452 mol
For O2:
moles = mass (grams) / molar mass (g/mol)
moles = 39.0g / (2 × 16.0 g/mol)
moles ≈ 1.22 mol
2. Determine the stoichiometric ratio.
From the balanced equation for the first reaction, we can see that 5 moles of O2 are required to react with 1 mole of Sb2S3. Therefore, the stoichiometric ratio is 5:1.
3. Use the stoichiometry to find the moles of O2 consumed.
Since we have a 5:1 stoichiometric ratio, the moles of O2 consumed would be:
moles of O2 consumed = 0.452 mol × 5 = 2.26 mol
4. Determine the moles of Sb formed.
From the balanced equation of the second reaction, we know that 2 moles of Sb are formed for every 4 moles of C consumed.
moles of Sb formed = 2.26 mol × 2 / 4 = 1.13 mol
5. Convert the moles of Sb formed to grams.
Using the molar mass of Sb (121.7 g/mol), we can calculate the mass:
grams = moles × molar mass
grams = 1.13 mol × 121.7 g/mol
grams ≈ 137.9 g
Therefore, approximately 137.9 grams of Sb are formed when 84.9g of Sb2S3 and 39.0g of O2 are used in the first reaction, and the product is used in the second reaction.