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xz_007 [3.2K]
3 years ago
12

Consider the reaction for the production of NO2 from NO:

Chemistry
1 answer:
telo118 [61]3 years ago
4 0
<span> so now we use the number of moles of NO 5.27 moles. use again the PV=nRT R=62.36367(L mmHg K−1   mol−1) T=35C + 273 = 308K P=632 mm Hg V=97.3 L n=632*97.3/62.3636367*308=3.20145770868 Look at the mole ratio of NO to NO2 its 2:2 they share the same number of moles. Atomic weight of each element in NO2 N:14 O:16 Molar Mass:(16*2)+(14*1)=46 g Actual number of moles=3.20 Theortical number of moles=5.20 Percentage Yield=(Actual Yield/Theortical Yield)*100</span>
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Because a water molecule has a negative end and a positive end, it displays_____?
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4 years ago
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When 16.3 g of magnesium and 4.52 g of oxygen gas react, how many grams of magnesium oxide will be formed? Identify the limiting
Tanzania [10]

Answer:

22.77 g.

he limiting reactant is O₂, and the excess reactant is Mg.

Explanation:

  • From the balanced reaction:

<em>Mg + 1/2O₂ → MgO,</em>

1.0 mole of Mg reacts with 0.5 mole of oxygen to produce 1.0 mole of MgO.

  • We need to calculate the no. of moles of (16.3 g) of Mg and (4.52 g) of oxygen:

no. of moles of Mg = mass/molar mass = (16.3 g)/(24.3 g/mol) = 0.6708 mol.

no. of moles of O₂ = mass/molar mass = (4.52 g)/(16.0 g/mol) = 0.2825 mol.

So. 0.565 mol of Mg reacts completely with (0.2825 mol) of O₂.

<em>∴ The limiting reactant is O₂, and the excess reactant is Mg (0.6708 - 0.565 = 0.1058 mol).</em>

<u><em>Using cross multiplication:</em></u>

1.0 mole of Mg produce → 1.0 mol of MgO.

∴ 0.565 mol of Mg produce → <em>0.565 mol of MgO.</em>

<em>∴ The amount of MgO produced = no. of moles x molar mass </em>= (0.565 mol)(40.3 g/mol) = <em>22.77 g.</em>

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3 years ago
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4 0
4 years ago
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What is the temperature of 0.5 moles of water vapor that occupies 120 dm3 and applies a pressure of 15,000 Pa to its container?
MArishka [77]
Use the ideal gas law:

PV = nRT

so, T = PV / nR

n=0.5
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Put the values:

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