# Question 027ec

Mar 9, 2015

You are correct about point a), the mass of the remaining uranium-238 is indeed 1.26 kg.

Now for point b). The trick here is to know that you don't get 1 gram of lead-206 for every 1 gram of uranium-238 that decays, but rather 1 mole of lead-206 for every 1 mole of uranium-238 that decays.

Since you're left with 1.26 kg, the mass of uranium that dcayed is

m_("decayed U-238") = "2.00 kg" - "1.26 kg" = "0.740 kg"

Convert this value to grams

$\text{0.740 kg" * "1000 g"/"1 kg" = "740. g}$

Now use uranium-238's molar mass to see how many moles decayed

$\text{740. g" * "1 mole"/"238.03 g" = "3.11 moles}$

This means of course that you've produced

$\text{3.11 moles U-238" * "1 mole Pb-206"/"1 mole U-238" = "3.11 moles Pb-206}$

This translates to a mass of

$\text{3.11 moles Pb-206" * "205.97 g"/"1 mole" * "1 kg"/"1000 g" = "0.6405 kg Pb-206}$

Rounded to three sig figs, the answer is

m_("Pb-206 produced") = "0.641 kg"#