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. How many molecules of N2 are produced when 15.9g of NO reacts with NH3?
4NH3 (g) + 6NO (g) → 5N2 (g) + 6H2O (g)


. How many molecules of N2 are produced when 15.9g of NO reacts with NH3?
4NH3 (g) + 6NO (g) → 5N2 (g) + 6H2O (g)

1 Answer

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To solve this problem, we need to first determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in a chemical reaction, limiting the amount of product that can be formed. To do this, we need to calculate the moles of each reactant present and compare their ratios.
Molar mass of NO (NO): 30.01 g/mol
Molar mass of NH3 (NH3): 17.03 g/mol

15.9 g of NO = 15.9 g / 30.01 g/mol = 0.530 moles of NO
Assuming excess NH3, moles of N2 produced = moles of NO consumed

From the balanced chemical equation, we see that 6 moles of NO produces 5 moles of N2.

6 moles of NO = 5 moles of N2
0.530 moles of NO = (5/6) * 0.530 = 0.4417 moles of N2

Now, we can use Avogadro's number to convert moles to molecules:

1 mole of any gas contains 6.022 x 10^23 molecules
0.4417 moles of N2 = 0.4417 mol x 6.022 x 10^23 molecules/mol = 2.657 x 10^23 molecules of N2

Therefore, 2.657 x 10^23 molecules of N2 are produced.

This is the same question as the first one, so the answer would be the same:
2.657 x 10^23 molecules of N2 are produced
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