Answer
5.93 g of Kr-91 is produced by the complete fission of 15 g of U-235.
Work Step by Step
Atomic mass of U-235 = 235.04 amu.
Hence, 1 mole U-235 = 235.04 g.
Therefore, 235.04 g of U-235 contain 6.023 $\times 10^{23}$ no. of U-235 atoms
15 g of U-235 will contain $\frac{6.023 \times 10^{23}\times15}{235.04}$ = 3.84 $\times10^{22}$ no. of U-235 atoms
These no. of atoms will produce same no. of Kr-91 atoms.
Atomic mass of Kr-91 = 92.93 g
hence, 6.023 $\times 10^{23}$ no. of Kr-91 have mass 92.93 g
Therefore, 3.84 $\times10^{22}$ no. of Kr-91 will have mass of $\frac{3.84 \times10^{22}\times92.93}{6.023\times 10^{23}}$ $\approx$ 5.93 g
Hence, the answer is 5.93 g.
