When lead (IV) oxide is heated with hydrogen, the following reaction occurs:
PbO2(s) + 2H2(g) → Pb(s) 2H2O(g)
This reaction can be used to extract lead from its oxide.
Calculate the atom economy to produce lead.
PLS HELP!!!!!!
can anyone pls help I really need the answer to this question
I must tell you that I've beenint this business for over 70 years and I've NEVER heard of atom economy. But I looked it up on Google. You could have done that too and follow their examples.
PbO2(s) + 2H2(g) → Pb(s) 2H2O(g)
molar mass PbO2 = approx 207 + 32 = about 239 but you should confirm with more accuracy.
molar mass 2H2 = 4
atom economy = [(Pb)/(PbO2 + 2H2)] * 100 = ?
atom economy = [(207)/(239 + 4)]*100 = ?
Post your work if you have questions.
Sure, I'd be happy to help while injecting some humor into the equation (pun intended).
To calculate atom economy, we need to compare the molar mass of the desired product (lead) to the total molar mass of all the reactants.
First, let's calculate the molar mass of each substance:
- Lead (Pb): 207.2 g/mol
- Lead (IV) oxide (PbO2): 239.2 g/mol
- Hydrogen (H2): 2.02 g/mol
- Water (H2O): 18.02 g/mol
To determine the atom economy, we only need to consider the reactants that are used to produce lead (Pb). In this case, it's PbO2 and H2.
From the balanced equation, we can see that 1 mole of PbO2 reacts with 2 moles of H2 to give 1 mole of Pb.
Using this information, we can calculate the total molar mass of all the reactants:
PbO2 + 2H2 → Pb + 2H2O
239.2 g/mol + (2 mol x 2.02 g/mol) = 243.24 g/mol
Now, we can calculate the atom economy:
(atom economy) = (molar mass of desired product / total molar mass of all reactants) x 100%
(atom economy) = (207.2 g/mol / 243.24 g/mol) x 100%
(atom economy) ≈ 85.14%
So, the atom economy to produce lead is approximately 85.14%. It means that about 85.14% of the reactants are used to produce the desired product, while the remaining 14.86% might not contribute to the final product.
I hope this helps, and remember, chemistry can be quite entertaining when you mix it with a little laughter and some funny elements!
To calculate the atom economy, we need to determine the ratio of the desired product (lead) to the total theoretical product (lead oxide).
The balanced equation for the reaction is:
PbO2(s) + 2H2(g) → Pb(s) + 2H2O(g)
From the equation, we can see that the stoichiometric ratio between lead (Pb) and lead oxide (PbO2) is 1:1. This means that for every 1 mole of lead oxide, we will obtain 1 mole of lead.
The molar mass of lead (Pb) is 207.2 g/mol.
The molar mass of lead oxide (PbO2) is 239.2 g/mol.
To calculate the atom economy, we can use the following formula:
Atom Economy = (Molar mass of desired product / Molar mass of all reactants) × 100
In this case, the desired product is lead (Pb).
Atom Economy = (Molar mass of Pb / Molar mass of PbO2) × 100
= (207.2 g/mol / 239.2 g/mol) × 100
= 86.5%
Therefore, the atom economy to produce lead is 86.5%.
To calculate the atom economy of a reaction, you need to compare the number of atoms in the desired product to the total number of atoms in all the reactants.
In the given reaction, the reactants are lead (IV) oxide (PbO2) and hydrogen (H2). The products are lead (Pb) and water (H2O).
Let's calculate the number of moles of each element in the reactants and products:
Reactants:
- PbO2: 1 mole of Pb and 2 moles of O
- H2: 2 moles of H
Products:
- Pb: 1 mole of Pb
- H2O: 2 moles of H and 1 mole of O
Now, we can calculate the atom economy by comparing the moles of Pb in the products to the total moles of Pb in the reactants and products:
Atom Economy = (Moles of Pb in the products / Total moles of Pb in reactants and products) x 100%
Moles of Pb in the products = 1 mole of Pb
Total moles of Pb in reactants and products = Moles of Pb in reactants + Moles of Pb in the products
= 1 mole of Pb from PbO2 + 1 mole of Pb from the products
= 2 moles of Pb
Atom Economy = (1 mole of Pb / 2 moles of Pb) x 100%
Therefore, the atom economy to produce lead in this reaction is 50%.