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Sagot :
89.6 moles of mno3 are produced when 4.3 kg of oxygen gas completely reacts according to the balanced chemical reaction.
How this is calculated?
- The balanced chemical reaction,
[tex]2Mn(s)+3O_{2} (g)- > 2MnO_{3} (s)[/tex]
- The molar ratio of the components in a chemical reaction is given by the balanced chemical equation. The moles of the reactants and products are indicated by the coefficients.
- Therefore, using a balanced chemical equation, it is feasible to represent the moles of one reactant in terms of the moles of another reactant or the moles of a product.
- When converting substances represented in grams to moles, the substance's molar mass is employed.
- using the oxygen's molar mass to convert the given kilograms of oxygen to grams and subsequently to moles.
4.30kg [tex]O_{2}[/tex] ×1000g [tex]O_{2}[/tex] /1kg [tex]O_{2}[/tex] ×1 mol [tex]O_{2}[/tex] /32.00g [tex]O_{2}[/tex] =134 mol [tex]O_{2}[/tex]
- Converting the given moles to moles of manganese (VI) oxide using the balanced chemical equation
[tex]134mol O_{2}[/tex] × [tex]2molMnO_{3}[/tex] / [tex]3mol O_{2}[/tex] =89.6 [tex]molMnO_{3}[/tex]
To know more about moles, refer:
https://brainly.com/question/24322641
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