Answer
$0.803\ kg/L$
$20,100\ L$
Work Step by Step
Converting the density:
$1.77\ lb/L\times\frac{453.6\ g}{1\ lb}\times \frac{1\ kg}{10^3\ g}=0.803\ kg/L$
Fuel required in L:
$22,300\ kg \div 0.803\ kg/L=27,800\ L$
Fuel in the tank: 7682 L
Fuel necessary to be added: $27,800-7,682=20,100\ L$
