Given 3.50 moles of hydrogen and 5.00 moles of nitrogen to produce ammonia, nitrogen is the excess reactant.
The excess reactant is the reactant in a chemical reaction with a greater amount than necessary to react completely with the limiting reactant.
3 H₂ + N₂ ⇒ 2 NH₃
The theoretical ratio (TR) of H₂ to N₂ is 3:1.
The experimental ratio (ER) of H₂ to N₂ is 3.50:5.00 = 0.70:1.
Comparing TR and ER, we can realize that there is not enough hydrogen to react with the nitrogen. Thus, nitrogen is the excess reactant.
Given 3.50 moles of hydrogen and 5.00 moles of nitrogen to produce ammonia, nitrogen is the excess reactant.
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Answer : The element is the reactant in excess.
Solution : Given,
Moles of = 5 moles
Moles of = 3.50 moles
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 3 moles of react with 1 mole of
So, 3.5 moles of react with moles of
The excess of = 5 - 1.16 = 3.84 moles
That means in the given balanced reaction, is a limiting reagent because it limits the formation of products and is an excess reagent.
Hence, the element is the reactant in excess.
To what element does the following electron configuration correspond?
1s22s22p4
A. aluminum: 13 electrons
B. oxygen: 8 electrons
C. boron: 5 electrons
Among the given figures, 2.5 x 10-2 L (or 25 mL when converted to milliliters) is the smallest volume.
The student is asked to compare and determine the smallest volume between 2500 mL, 250 cm3, 2.5 x 10-2 L, and 25 m3. First, let's convert all volumes into a common unit, the milliliter (mL).
From this comparison, we can see that 2.5 x 10-2 L or 25 mL is the smallest volume among the given figures.
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The mass percent of sodium in NaCl can be determined by dividing the mass of sodium by the total mass of the compound (NaCl), then multiplying the result by 100. So, 39.34% of the mass of sodium in NaCl is composed of sodium.
The molecular weight of chlorine (Cl) is 35.45 g/mol while that of sodium (Na) is 22.99 g/mol.
The molecular weight of NaCl is calculated as follows:
Total molecular weight = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol
To determine the sodium mass percent in NaCl:
(mass of sodium / total mass of NaCl) * 100 to get the mass percent of sodium.
(22.99 g/mol / 58.44 g/mol) x 100 = 39.34% sodium mass percent
As a result, 39.34% of the mass of sodium in NaCl is composed of sodium.
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