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How many grams of nitrogen gas are produced when 48.00 moles of Sodium Hydroxide reacts with Nitrogen Tribromide?

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Answer:

448g

Step-by-step explanation:

First, we'll begin by writing the balanced equation for the reaction. This is illustrated below:

3NaOH + 2NBr3 → N2 + 3BrNa + 3HBrO

Next, we shall determine the number of mole of N2 produce by the reaction of 48 moles of NaOH.

This is illustrated below:

From the balanced equation above, 3 moles of NaOH produced 1 mole of N2.

Therefore, 48 moles of NaOH will produce = 48/3 = 16 moles of N2

Finally, we shall convert 16 moles of N2 to grams. This can be achieved as shown below:

Number of mole of N2 = 16 moles

Molar Mass of N2 = 2x14 = 28g/mol

Mass of N2 =?

Mass = number of mole x molar Mass

Mass of N2 = 16 x 28

Mass of N2 = 448g.

Therefore, 448g of N2 is produced from the reaction.

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