nabr + cacl2 are the products
The rock cycle
The rock cycle can have sedimentary, igneous, or metamorphic rocks going back and forth changing into different rocks.
No of atoms=No of moles ×Avagadro no
![\\ \tt\longmapsto 4\times 6.023\times 10^{23}](https://tex.z-dn.net/?f=%5C%5C%20%5Ctt%5Clongmapsto%204%5Ctimes%206.023%5Ctimes%2010%5E%7B23%7D)
![\\ \tt\longmapsto 24.092\times 10^{23}](https://tex.z-dn.net/?f=%5C%5C%20%5Ctt%5Clongmapsto%2024.092%5Ctimes%2010%5E%7B23%7D)
![\\ \tt\longmapsto 2.4\times 10^{24}atoms](https://tex.z-dn.net/?f=%5C%5C%20%5Ctt%5Clongmapsto%202.4%5Ctimes%2010%5E%7B24%7Datoms)
Report this clown who put the first answer he’s trying to get your ip
<h3>Answer:</h3>
6 moles of H₂O / 9 moles of O₂
<h3>Explanation:</h3>
The Balance Chemical Equation is as follow,
2 C₃H₆ + 9 O₂ → 6 CO₂ + 6 H₂O
According to this balance equation 2 moles of C₃H₆ (Propene) when combusted with 9 moles of O₂ (Oxygen) produces 6 moles of CO₂ (Carbon Dioxide) and 6 moles of H₂O (Water Vapors).
Hence, Given options are;
1 mole of C₃H₆ / 2 moles of CO₂: This is incorrect because 1 mole of C₃H₆ will produce 2 moles of CO₂.
6 moles of H₂O / 9 moles of O₂: This is correct as 6 moles of H₂O is produced by reacting 9 moles of O₂.
2 moles of C₃H₆ / 6 moles of O₂: This ratio is incorrect because for complete oxidation 2 moles of C₃H₆ requires 9 moles of O₂.
3 moles of H₂O / 2 moles of CO₂: This is also incorrect because the production of 3 moles of H₂O will be accompanied by the production of 3 moles of CO₂.