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11.2 L of oxygen STP and 8 grams of calc...

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

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The balanced chemical equation is `2Ca+O_(2)rarr2CaO`
2 moles of Ca = 1 mole of `O_(2)`
`(2xx40)` grams of Ca = 22.4L of `O_(2)` at STP
8 grams of Ca = 2.24 L of `O_(2)` at STP
The limiting reagent is calcium.
Volume of `O_(2)` at STP required = 2.24L
Volume of `O_(2)` left unreacted = 11.2-2.24 = 8.96L at STP
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