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lbvjy [14]
3 years ago
14

37.5g of nitrogen reacts with 15.5 g of hydrogen. What mass of ammonia can be made? What is the limiting reactant?

Chemistry
1 answer:
Gennadij [26K]3 years ago
7 0

Answer:

Limiting reactants or limiting reagents decide the amount of product formed and the amount of excess reagent used.

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Which is a component of John Dalton’s atomic theory?
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A cough sytup contains 0.5M dextromethophan. How many moles of the cough supressant are in 21.3mL of the cough syrup?
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Answer:

0.0107 mol

Explanation:

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3 years ago
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11111nata11111 [884]

Formula mass, molar mass and Avagadro's number.

Explanation:

number of atoms in a compound can be calculated by knowing the molar mass of the compound or element, the result will be multiplied by avagadro's number (6.022*10^23)

1 mole of a substance is equal to Avagadro number of atoms.

If the number of moles is known of a compound or element its molar mass can be calculated as:

n= Weight of the compound/element given/ molecular weight of the same.

  formula mass is the mass of compound ie chemical compound formed with different molecules. its mass is calculated by adding the molar masses of all the elements taking part in its assembly.

5 0
3 years ago
In 1909 Fritz Haber discovered the workable conditions under which nitrogen, N2(g), and hydrogen, H2(g), would combine using to
labwork [276]

Answer : 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

Solution : Given,

Mass of NH_3 = 100 g

Molar mass of NH_3 = 27 g/mole

Molar mass of N_2 = 28 g/mole

First we have to calculate moles of NH_3.

\text{ Moles of }NH_3=\frac{\text{ Mass of }NH_3}{\text{ Molar mass of }NH_3}= \frac{100g}{27g/mole}=3.7moles

The given balanced chemical reaction is,

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

From the given reaction, we conclude that

2 moles of NH_3 produced from 1 mole of N_2

3.7 moles of NH_3 produced from \frac{1mole}{2mole}\times 3.7mole=1.85moles of N_2

Now we have to calculate the mass of N_2.

Mass of N_2 = Moles of N_2 × Molar mass of N_2

Mass of N_2 = 1.85 mole × 28 g/mole = 51.8 g

Therefore, 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

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3 years ago
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