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How many grams of O₂ are required to react completely with 14.6 g of Na to form sodium oxide, Na₂O?

User Rudy
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2 Answers

3 votes

Final answer:

To form sodium oxide, Na₂O, from 14.6 g of Na, you need 11.3 g of O₂.

Step-by-step explanation:

To determine how many grams of O₂ are required to react completely with 14.6 g of Na to form sodium oxide, Na₂O, we need to use the balanced chemical equation and the concept of stoichiometry. The balanced equation is:

4Na(s) + O₂(g) ➝ 2Na₂O(s)

From the equation, we can see that 4 moles of Na reacts with 1 mole of O₂ to form 2 moles of Na₂O. We can set up a ratio using the molar masses:

(14.6 g Na) × (1 mol Na/22.99 g Na) × (1 mol O₂/4 mol Na) × (32.00 g O₂/1 mol O₂) = 11.3 g O₂

Therefore, 11.3 grams of O₂ are required to react completely with 14.6 g of Na.

User Diana Mikhasyova
by
7.9k points
4 votes

The balanced chemical reaction is :


O_2 + 4Na \ -> \ 2Na_2O

Number of moles of Na,
n = (14.6)/(23) = 0.635 \ mol .

Now, from balance chemical reaction we can see that 1 mole of oxygen reacts with 4 moles of sodium.

So, number of moles of oxygen are :


n = (0.635)/(4)\ mole

So, amount of oxygen required is :


m = (0.635 * 32)/(4)\ gm\\\\m = 5.08 \ gm

Therefore, 5.08 gram of oxygen will react with 14.6 gram of sodium.

User John Rood
by
7.9k points
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