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Tomtit [17]
3 years ago
5

If 25.0 g of NH3 and 38.9g of O2 react in the following reaction, how

Chemistry
1 answer:
Elan Coil [88]3 years ago
8 0

Answer:

There will be formed 36.49 grams of NO

Explanation:

<u>Step 1</u>: The balanced equation

4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H20 (g)

<u>Step 2:</u> Data given

Mass of NH3 = 25.0 grams

Mass of O2 = 38.9 grams

Molar mass of NH3 = 17.03 g/mol

Molar mass of 02 = 32 g/mol

Molar mass of NO = 30.01 g/mol

<u>Step 3:</u> Calculate number of moles

Moles of NH3 = 25.0 grams / 17.03 g/mol

Moles of NH3 = 1.47 moles

Moles of O2 = 38.9 grams / 32 g/mol

Moles of O2 = 1.216 moles

<u>Step 4:</u> Calculate limiting reactant

For 4 moles of NH3 consumed, we need 5 moles of O2 to produce 4 moles of NO and 6 moles of H2O

O2 is the limiting reactant. It will complete be consumed (1.216 moles).

NH3 is the limiting reactant. There will 4/5 * 1.216 = 0.9728 moles of NH3 consumed.

There will remain 0.4972 moles of NH3.

<u>Step 5: </u>Calculates moles of NO

If there is consumed 4 moles of NH3, there is produced 4 moles of NO

For 1.216 moles of NH3 consumed, we will have 1.216 moles of NO

<u>Step 6:</u> Calculate mass of NO

Mass NO = moles NO * Molar mass NO

Mass NO = 1.216 moles * 30.01 g/mol = 36.49 grams

There will be formed 36.49 grams of NO

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Allushta [10]

Answer:

<h2><u>Reason:</u></h2>

Catalyst is used as a very fine powder and larger pieces of iron are not used. This is because the surface area of catalyst needs to be large so that more of the surface is exposed to the substrate and more of the substrate is catalyzed.

<h2><u>Important Info:</u></h2>

=> Larger Pieces of Iron has a smaller surface area than the fine particles.

=> Larger the surface area of catalysts/enzymes , more will be the reaction rate and vice versa.

Hope this helped!

<h2>~AnonymousHelper1807</h2>

7 0
3 years ago
How many kilowatt-hours of electricity are used to produce 4.50 kg of magnesium in the electrolysis of molten mgcl2 with an appl
ratelena [41]
First, we need to determine the half reaction of magnesium. It would be expressed as:

Mg2+ + 2e- = Mg

Given the mass of magnesium metal that is produced, we calculate the total charge of the electrolysis by the relations we can get from the half reaction. We do as follows:

4.50 kg Mg ( 1000 g / 1 kg ) ( 1 mol / 24.305 g ) ( 2 mol e- / 1 mol Mg ) ( 96500 C / 1 mol e- ) = 35733388.2 C

We are given the applied EMF in units of V. This value is equal to J/C. So, 5 V is equal to 5 J/C.

35733388.2 C (5 J/C) = 178666941 J
178666941 J ( 1 kW-h / 3.6x10^6 J ) = 49.63 kW-h
3 0
3 years ago
How many moles of ammonia (nh3) would be produced if 2.5 moles of nitrogen (n2) reacted with excess hydrogen (h2)?
cluponka [151]
Firstly, a balanced equation has to be written for the production of ammonia (NH₃) from hydrogen gas (H₂) and nitrogen gas (N₂):
                    N₂   +   3H₂   →   2NH₃

Now, the mole ratio of N₂ : NH₃ is 1 : 2 based on the coefficients of the balanced equation.

If the moles of N₂ = 2.5 moles

   then the moles of NH₃ produced = 2.5 mol × 2 
                                                          =  5 mol

Thus, the moles of ammonia produced when 2.5 mol of nitrogen gas is combined with excess hydrogen gas is 5 mol.
6 0
3 years ago
12. Study the chemical reaction below, what are the product(s) in the reaction and what are the reactant(s) in the reaction (Sho
natita [175]

Explanation:

The reactants are Aluminium and Copper chloride.

The products are Copper and Aluminium chloride.

5 0
3 years ago
2NBr3 + 3NaOH = N2 +3NaBr + 3HOBr
bija089 [108]
The theoretical proportion is given by the balanced chemical equation:

2 mol NBr / 3 mol  Na OH

Then x mol NaOH / 40 mol NBr3 = 3mol NaOH/2 mol NBr3

Solve for x, x = 40 * 3/2 = 60 mol NaOH.

Given that there are 48 mol NaOH (less than 60) this is the limitant reactant and the other is the excess reactant.

Answer: NBr3..

 
6 0
3 years ago
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