Sodium azide (NaN3) yields N2 gas when heated to 300 °C, a reaction used in automobile air bags. If 1.00 mol of N2 has a volume of 47.0 L under the reaction conditions, how many liters of gas can be formed by heating 38.5 g of NaN3? The reaction is

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well, everyone in this video are being asked how much gas and Leaders is produced when we have 24.7 g of H two C 03 that's heated, let's first of all write the equation for this. So of course we have certain material, H two C 03 and we're adding heat to this. What we'll get is H two Oh, of course that's in its liquid state. And carbon dioxide, which is a gas. So we can see here while overlooking for how much gas. And Leaders, the only source of gas is going to be our carbon dioxide. So let's first start off our calculations with the grounds of our story materials that's given to us being 24. g of H two C 03. We use the molar mass of this to give us two moles of H two CO 03. So for every one mole of H two C 03, we have 62.3 g than using the multi mole ratio, I guess, to the most of Parliament dioxide. So for every one mole of carbon dioxide we have one mole of H two C 03. And then finally using the volume of carbon dioxide, we get that for everyone more of our CO2, There is 32.85 Leaders of CO two. So you can see here that the grams of H two, C 030 cancel moles of H two C three, it will cancel and moles of C 02 cancel. Now, if we put everything into the calculator will get the numerical value of 13.8 and units, of course, being leaders of carbon dioxide. And that's just going to be the amount of gas and leaders that's produced when we have 24.7 g of H two c 03 that's being heated. That's going to be my final answer, and thank you all so much for watching.

Sodium azide (NaN3) yields N2 gas when heated to 300 °C, a reaction used in automobile air bags. If 1.00 mol of N2 has a volume of 47.0 L under the reaction conditions, how many liters of gas can be formed by heating 38.5 g of NaN3? The reaction is

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Key Concepts

Stoichiometry

Molar Mass

Ideal Gas Law