Molar mass of N = 14 g/molMolar mass of O2 = 32 g/molAdding both masses = 46 g/molActual molar mass/ Empirical molar mass = 138.02 / 46 = 3Now multiplying this co effecient with empirical fomula NO2 = 3(NO2) = N3O6So according to above explanation,D) N3O6, is the correct answer.
Answer:

Explanation:
Hello,
In this case, the described chemical reaction is:

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

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Answer:
40.8g of sodium sulfate must be added
Explanation:
The reaction of barium nitrate, Ba(NO₃)₂ with sodium sulfate, Na₂SO₄ is:
Ba(NO₃)₂ + Na₂SO₄ → 2 NaNO₃ + BaSO₄(s)
That means, for a complete reaction of an amount of barium nitrate you must add the same amount in moles of sodium sulfate. To solve this problem we need to convert the mass of barium nitrate to moles = Moles of sodium sulfate that must be added:
<em>Moles Ba(NO₃)₂ -Molar mass: 261.3g/mol-:</em>
75g * (1mol / 261.3g) = 0.287 moles = Moles Na₂SO₄
<em>Mass Na₂SO₄ -Molar mass: 142.04g/mol-:</em>
0.287 moles * (142.04g / mol) =
<h3>40.8g of sodium sulfate must be added</h3>