For the reaction, 4P4 + 5S8 ==> 4P4S10, what is the maximum amount of P4S10 which could be formed from 6.69 mol of P4 and 14.61 mol of S8?

Balance equation:   4P4  +  5S8  ==⇒  4P4S10

13.19 g P4  x 1mole P4/123.9 g x  4 moles P4S10/4 moles P4 = 0.106 moles P4S10

1.88 g S8  x  1mole S8/256.5 g x 4 moles P4S10/5moles S8 = 0.00586 moles P4S10

One cannot obtain more than 0.00586 moles of P4S10.  So, the limiting reactant is S8.

0.00586 moles P4S10  x  445 g P4S10/1mole P4S10 = 2.61 g P4S10

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.