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2gram of H2 react with 8gram of O2 . Calculate the maximum amount of O2 formed.?
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2gram of H2 react with 8gram of O2 . Calculate the maximum amount of O...



Reaction:
- The reaction between 2g of H2 and 8g of O2 can be represented as:
2H2 + O2 -> 2H2O

Calculations:
- Firstly, we need to determine the limiting reactant in the reaction.
- Calculate the moles of each reactant using their molar masses:
- Moles of H2 = 2g / 2g/mol = 1 mol
- Moles of O2 = 8g / 32g/mol = 0.25 mol
- From the balanced chemical equation, we see that 1 mol of O2 reacts with 2 mol of H2, so the given amount of O2 is in excess.
- Therefore, H2 is the limiting reactant.
- Now, calculate the maximum amount of O2 formed by using the moles of H2 reacted:
- Moles of H2 = 1 mol
- Moles of O2 produced = 1 mol H2 * (1 mol O2 / 2 mol H2) = 0.5 mol
- Convert moles of O2 to grams:
- Mass of O2 = 0.5 mol * 32g/mol = 16g

Maximum Amount of O2 Formed:
- The maximum amount of O2 formed in the reaction is 16 grams.
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2gram of H2 react with 8gram of O2 . Calculate the maximum amount of O2 formed.?
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