Question

5g nitrogen and 12g hydrogen react to form Ammonia . which is the limiting reagent in the reaction​

Answer 1

Answer:

Thus for the given case N2 will be the limiting reagent.

Explanation:

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Answer 2

Answer:

The limiting reagent in this reaction is nitrogen.

Explanation:

The balanced chemical equation for the reaction can be written as:

$N_{2} +3H_{2}$ → $2NH_{3}$

Given mass of nitrogen = 5 g

Molar mass of nitrogen = (14×2) = 28 g/mol

Number of moles of nitrogen = given mass ÷ molar mass

                                                 = $\frac{5}{28}$ = 0.18 moles

Given mass of hydrogen = 12 g

Molar mass of hydrogen = (1×2) = 2 g/mol

Number of moles of hydrogen = given mass ÷ molar mass

                                                   = $\frac{12}2}$ = 6 moles

According to the balanced chemical equation,

1 mole of nitrogen reacts with 2 moles of hydrogen to give 2 moles of ammonia.

Here we have 0.18 moles of nitrogen.

∴Number of moles of hydrogen required = 0.18 × 3 = 0.54 moles

But we have 6 moles of hydrogen which is excess.

Therefore the limiting reagent is nitrogen.