The u. s. mint produces a dollar coin called the american silver eagle that is made of nearly pure silver. this coin has a diameter of 41 mm and a thickness of 2.5 mm. the density and approximate market price of silver are 10.5 g/cm3 and $0.51 per gram, respectively. calculate the value of the silver in the coin, assuming its thickness is uniform.

The value of the silver in the coin is 35.3 $

Explanation:

First of all, let's calculate the volume of the coin.

2π . r² . thickness = volume

r = diameter/2

r = 41 mm/2 = 20.5 mm

2 . π . (20.5 mm)² .  2.5 mm = 6601 mm³

Now, this is the volume of the coin, so we must find out how many grams are on it.

6601 mm³ / 1000 = 6.60 cm³

Let's apply density.

D = Mass / volume

10.5 g/cm³ = mass /6.60 cm³

10.5 g/cm³ . 6.60 cm³ = mass

69.3 g = mass

Each gram has a cost of 0.51$

69.3 g . 0.51$ = 35.3 $

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2