A major motor company displays a die-cast model of its first automobile, made from 9.35kg of iron. To celebrate its one-hundreth year in business, a worker will recast the model in gold from the original dies. What mass of gold is needed to make the new model?
~my thoughts:
m= 9.35kg
atomic mass of gold= 196.97g/mol
atomic mass of iron= 55.845g/mol
9.35kg(1000g/1kg)(1molFe/55.845g)(196.97g/1mol Au)(1kg/1000g)= 33.0kg of Au
~The answer in the book is 23.0kg so I don't know what I did wrong, however I think that the problem is btwn the moles of Fe and moles of Au conversion b/c usually in chem you have a rxn where you say the ratio of the 2 compounds to each other.
I'm not sure where I went wrong
can someone help me out here?
~Thanks~
Thank you drwls, I got the same answer as the book =D
ρ iron = 7870 kg/m3 m iron =9.35kg
ρ gold =19320 kg/m3
v=m/ρ ρ=m/v
v=9.35/7870
v=1.1880×10^(-3)
m=v×ρ m=1.1880×10^(-3)×19320
m=22.9 kg≈23kg
To calculate the mass of gold needed to make the new model, you need to convert the mass of iron in the original model into moles, and then use the molar ratio between iron and gold to determine the mass of gold required.
Here is the correct calculation:
Given:
Mass of iron in the original model (m) = 9.35 kg
Atomic mass of gold (Au) = 196.97 g/mol
Atomic mass of iron (Fe) = 55.845 g/mol
Step 1: Convert the mass of iron into grams
9.35 kg x 1000 g/kg = 9350 g
Step 2: Convert the mass of iron into moles
9350 g / 55.845 g/mol = 167.13 mol Fe
Step 3: Use the molar ratio between iron and gold to find the mass of gold
Using the balanced equation of the reaction between iron and gold, which we can assume to be 1:1 (since the same model is being recast), we have:
1 mol Fe : 1 mol Au
Therefore,
167.13 mol Fe : x mol Au
x = 167.13 mol
Step 4: Convert the moles of gold into grams
167.13 mol x 196.97 g/mol = 32,913.44 g
Step 5: Convert the mass of gold into kilograms
32,913.44 g / 1000 = 32.91 kg
So, according to the calculations, the mass of gold needed to make the new model is approximately 32.91 kg.
Since the answer in the book is given as 23.0 kg, it is possible that there may be an error in the given atomic masses or another aspect of the question. I would recommend double-checking your values and discussing the problem with your teacher or classmates to clarify any discrepancies.