11.2 L of oxygen STP and 8 grams of calcium are allowed to react. What volume of which chemical 3 is left unreacted? 

Explanation

The balanced chemical equation is
`2Ca+O_(2)rarr 2CaO`
2 moles of `Ca -= 1` mole of `O_(2)`
`(2 xx 40)` grams of `Ca = 22.4 L` of `O_(2)` at STP
8 grams of Ca = 2.24 L of `O_(2)` at STP
40 grams of Ca = 11.2 L of `O_(2)` at STP
The limiting reagent is calcium.
Volume of `O_(2)` at STP required = 2.24 L
Volume of `O_(2)` left unreacted `= 11.2 - 2.24 = 8.96 L` at STP