Question

3 Cu + 8HNO3 g 3 Cu(NO3)2 + 2 NO + 4 H2O

Answer 1

24.25 moles of NO would be made when 97 moles of $HNO_{3}$ are consumed

Given reaction:

$3 Cu + 8HNO_{3} - > 3 Cu(NO_{3})_{2} + 2 NO + 4 H_{2}O$

As the given reaction is already balanced so we don't need to balance it

In a balanced chemical reaction the ratio of reactants consumed and products formed always remains constant.

∵ In the given Reaction we can analyse for 8 moles $HNO_{3}$ consumed 2 moles of $NO$ are being produced.

Assuming $x$ moles of $NO$ being produced when $97$ moles $HNO_{3$ produced

∵ Ratio remains constant

⇒$\frac{HNO_{3} Consumed}{NO Produced} = \frac{8}{2} =\frac{97}{x}$

⇒$x=\frac{97}{4}=24.25$ moles

∴ 24.25 moles of $NO$ would be made.

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Answer 2

24.25 moles of NO can be produced using 97 moles of HNO3.

What is balanced chemical equation?

Equal numbers of atoms from various elements are present in both the reactants and the products in balanced chemical equations. Varied elements' atom counts in the reactants and products of unbalanced chemical equations are different.

3 Cu + 8HNO3 g → 3 Cu(NO3)2 + 2 NO + 4 H2O

The number of moles consumed can be calculated using comparing with coefficients in the balanced reaction .

So , from above eq we get that 8 moles of HNO3 are consumed to make 2 moles of NO.

⇒  8 HNO3⇔2 NO

⇒ 1 HNO3⇔ 1/4 NO

This means that for each mole of HNO3 produces 1/4 moles of NO.

So , for 97 moles of HNO3 , $\frac{1}{4} *97$ moles of NO can be made,

So, total moles of NO made are 24.25 moles.

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