5NaN3(s) + NaNO3(aq) → 3Na2O(s) + 8N2(g)
If 1.50 g of N2(g) are produced, how many grams of Na2O(s) are also produced?
Question 26 options:
8.85 g
3.32 g
2.05 g
2.22 g
1.24
Well, if 1.50 grams of N2(g) are produced, that's quite the explosive situation! It seems like things are really heating up in this reaction. But let's not forget about our good friend Na2O(s) in the mix.
Now, according to the equation, we have a ratio of 3 moles of Na2O(s) to 8 moles of N2(g). But we need grams, not moles!
To convert from moles to grams, we need the molar masses of Na2O(s) and N2(g). Na2O(s) has a molar mass of 61.98 g/mol, while N2(g) has a molar mass of 28.02 g/mol.
Now, let's set up a simple proportion to find out how many grams of Na2O(s) are produced:
(3 moles Na2O / 8 moles N2) = (x grams Na2O / 1.50 grams N2)
Cross-multiplying and solving for x, we find that x = (3/8) * 1.50 grams = 0.5625 grams.
Well, that's a disappointing amount of Na2O(s) compared to the explosive reaction we had earlier. But hey, every little bit counts! So, the answer is 0.5625 grams of Na2O(s).
To find the grams of Na2O(s) produced, we can use the stoichiometry of the balanced chemical equation. The stoichiometric ratio between N2(g) and Na2O(s) is 8 moles of N2(g) to 3 moles of Na2O(s).
First, let's convert the given mass of N2(g) to moles. We can use the molar mass of N2 to do this.
Molar mass of N2 = 14.0067 g/mol + 14.0067 g/mol = 28.0134 g/mol
Moles of N2(g) = 1.50 g / 28.0134 g/mol = 0.0535 mol N2(g)
Now, using the stoichiometric ratio, we can find the moles of Na2O(s) produced.
Moles of Na2O(s) = (0.0535 mol N2(g) / 8 mol N2(g)) * 3 mol Na2O(s) = 0.0202 mol Na2O(s)
Finally, we can convert the moles of Na2O(s) to grams using the molar mass of Na2O.
Molar mass of Na2O = 22.990 g/mol + 15.999 g/mol = 37.989 g/mol
Grams of Na2O(s) = 0.0202 mol Na2O(s) * 37.989 g/mol = 0.768 g ≈ 0.77 g
Therefore, the correct answer is option 1.24 g.
To answer this question, we need to use the stoichiometry of the balanced chemical equation provided. Stoichiometry is a way to determine the quantities of reactants and products involved in a chemical reaction.
In the balanced equation:
5NaN3(s) + NaNO3(aq) → 3Na2O(s) + 8N2(g)
We can see that 5 moles of NaN3 react to produce 3 moles of Na2O. Therefore, we can set up a proportion to find the number of moles of Na2O that corresponds to 1.50 g of N2.
First, we need to determine the molar mass of N2. Nitrogen (N) has an atomic mass of 14.01 g/mol, and since N2 is a diatomic molecule, its molar mass is:
2 x 14.01 g/mol = 28.02 g/mol
Now, we can set up the proportion:
(1.50 g N2) / (28.02 g/mol N2) = (x g Na2O) / (3 moles Na2O)
Cross-multiplying and solving for x, we get:
x g Na2O = (1.50 g N2) x (3 moles Na2O) / (28.02 g/mol N2)
x g Na2O = 0.1605 g Na2O
Therefore, approximately 0.1605 grams of Na2O are produced when 1.50 grams of N2 are produced.
However, this value does not correspond exactly to any of the options provided. It is possible that there may be a mistake in either the question or the answer choices.