1) Chemical equation
16Fe(s) + 3S8(s) ---> 8Fe2S3
2) Molar ratios:
16 mol Fe : 3 mole S8 : 8 mol Fe2S3
3) Convert masses in grams to number of moles
number of moles = mass in grams / molar mass
a) iron, Fe
mass = 3.0 g
atomic mass = 55.845 g/mol
=> number of moles of Fe = 3.0g / 55.845 g/mol = 0.0537 mol
b) Sulfur, S8
mass = 2.5 g
molar mass = 8*32.065 g/mol = 256.52 g/mol
=> number of moles of S8 = 2.5g / 256.52 g/mol = 0.009746 mol
4) Limiting reactant
Theoretical ratio actual ratio
16 mol Fe / 3 mol S8 0.0537 mol Fe / 0.009746 mol S8
5.33 5.50
So, there is a little bit more Fe than the theoretical needed to react all the S8, which means the S8 is the limiting reactant.
5) Calculate the number of moles of iron (III) produced with 2.5 g (0.009746 moles) of S8
3moles S8 / 8 moles Fe2S3 = 0.009746 moles S8 / x
=> x = 0.009746 * 8 / 3 moles Fe2S3 = 0.026 moles Fe2S3
6) Convert 0.026 moles Fe2S3 into grams
mass in grams = number of moles * molar mass
molar mass of Fe2S3 = 207.9 g/mol
mass = 0.026 mol * 207.9 g/mol = 5.40 g
7) Answer: option D)
Explanation:
1) Initial mass of the Cesium-137=
= 180 mg
Mass of Cesium after time t = N
Formula used :
Half life of the cesium-137 =
= initial mass of isotope
N = mass of the parent isotope left after the time, (t)
= half life of the isotope
= rate constant

Now put all the given values in this formula, we get
Mass that remains after t years.

Therefore, the parent isotope remain after one half life will be, 100 grams.
2)
t = 70 years


N = 35.73 mg
35.73 mg of cesium-137 will remain after 70 years.
3)


N = 1 mg
t = ?

t = 224.80 years ≈ 225 years
After 225 years only 1 mg of cesium-137 will remain.
<span>35.0 mL of 0.210 M
KOH
molarity = moles/volume
find moles of OH
do the same thing for: 50.0 mL of 0.210 M HClO(aq) but for H+
they will cancel out: H+ + OH- -> H2O
but you'll have some left over,
pH=-log[H+]
pOH
=-log[OH-]
pH+pOH
=14</span>