## Chemistry and Chemical Reactivity (9th Edition)

$0.803\ kg/L$ $20,100\ L$
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: $27,800-7,682=20,100\ L$