Respuesta :
Answer: 448 g of [tex]O_2[/tex] will be required to completely react with 784g moles of CO(g) during this reaction.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} CO=\frac{784g}{28g/mol}=28moles[/tex]
The balanced chemical equation is:
[tex]2CO(g)+O_2(g)\rightarrow 2CO_2(g)[/tex]
According to stoichiometry :
2 moles of [tex]CO[/tex] require = 1 mole of [tex]O_2[/tex]
Thus 28 moles of [tex]CO[/tex] will require=[tex]\frac{1}{2}\times 28=14moles[/tex] of [tex]O_2[/tex]
Mass of [tex]O_2=moles\times {\text {Molar mass}}=14moles\times 32g/mol=448g[/tex]
Thus 448g of [tex]O_2[/tex] will be required to completely react with 784g moles of CO(g) during this reaction.
Answer:
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Explanation:
