Considering the reaction of hydrogen and oxygen to produce water. If 24.6 g of water is produced and 2.9 g of hydrogen is used, the amount of oxygen needed is 21.87 grams
The best way to solve this question is to write out the reaction between hydrogen and oxygen that resulted in the formation of water.
So, we have:
[tex]\mathbf{ 2 H_2{(g)} + O_2{(g)} \to 2H_2O_{(l)}}[/tex]
From the above reaction;
2 moles of hydrogen gas reacted with 1 mole of oxygen to produce 2 moles of water.
If;
- 1 mole of O₂ = 32 grams of O₂
- 2 mole of H₂O = 2 × 18 grams of H₂O
- 2 mole of H₂O = 36 grams of H₂O
We can deduce that since 1 mole of O₂ produced 2 moles of H₂O;
Thus, 32 grams of O₂ is needed to produce 36 grams of H₂O.
∴
To produce, 24.6 grams of H₂O, we have:
[tex]\mathbf{=\Big(\dfrac{32 \times 24.6 }{36}\Big) grams }[/tex]
= 21.87 grams of oxygen.
Therefore, we can conclude that the amount of oxygen needed is 21.87 grams.
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