How many moles of H2 are needed to react with 1 mole off N2?

Answers

Answer 1
Answer: 3 H + N = 2 NH

3 moles H₂ ------- 1 mole N₂
? moles H₂ -------- 1 mole N₂

H₂ = 1 * 3 / 1

H₂ = 3 / 1

= 3 moles of H₂

hope this helps!

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Ammonia and sulfuric acid react to form ammonium sulfate. Determine the starting mass of each reactant if 20.3 g of ammonium sulfate is produced and 5.89g of sulfuric acid remains unreacted.

Answers

Answer: The mass of ammonia is 5.236 g and that of sulfuric acid is 15.064 g

Explanation:

  • Calculating the mass of ammonia:

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

Given mass of ammonium sulfate = 20.3 g

Molar mass of ammonium sulfate = 132.14 g/mol

Putting values in equation 1, we get:

\text{Moles of ammonium sulfate}=(20.3g)/(132.14g/mol)=0.154mol

The chemical equation for the reaction of ammonia and sulfuric acid follows:

2NH_3+H_2SO_4\rightarrow (NH_4)_2SO_4

As, sulfuric acid remains unreacted, which means it is an excess reagent and its starting mass cannot be determined from ammonium sulfate.

By Stoichiometry of the reaction:

1 mole of ammonium sulfate is produced by 2 moles of ammonia.

So, 0.154 moles of ammonium sulfate is produced by = (2)/(1)* 0.154=0.308mol of ammonia.

Now, calculating the mass of ammonia from equation 1, we get:

Molar mass of ammonia = 17 g/mol

Moles of ammonia = 0.308 moles

Putting values in equation 1, we get:

0.308mol=\frac{\text{Mass of ammonia}}{17g/mol}\n\n\text{Mass of ammonia}=5.236g

  • Calculating the mass of sulfuric acid

Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.  

This also means that total mass on the reactant side must be equal to the total mass on the product side.

Let the mass of sulfuric acid be 'x' grams

We are given:

Mass of ammonium sulfate = 20.3 grams

Mass of ammonia = 5.236 grams

Total mass on reactant side = 5.236 + x

Total mass on product side = 20.3 g

So, by applying law of conservation of mass, we get:

5.236+x=20.3\n\nx=15.064g

Hence, the mass of ammonia is 5.236 g and that of sulfuric acid is 15.064 g

2NH₃    +    H₂SO₄     ⇒     (NH₄)₂SO₄
34.06g  :     98.08g      :      132.14g
     x               y                 20.03g

x = [20.03g*34.06g]/132.14g = 5.16g

5.16g + y = 20.03g  ⇔  from stoichiometric ratio
20.03g - 5.16g = y
y = 14,87g + 5.89g = 20.76g  
⇔ 5.89g of sulfuric(VI) acid remains unreacted

Starting mass:

mNH₃ = 5.16g
mH₂SO₄ = 20.76g

Which ion is part of every amino acid?

Answers

In an amino acid, there are two ions present - the positively charged and the negatively charged. The positively charged is called the amino group while the negatively charged is the carboxyl group.

In plants, movement of CO2 and O2 happens through:

Answers

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Stomata are the pores present on the leaves of the plant

Which particles are gained and lost during a redox reaction?(1) electrons (3) neutrons
(2) protons (4) positrons

Answers

Electrons are Loss (Oxidation) and Gain (Reduction) in a Reduction-Oxidation (Redox) reaction.

If energy can neither be created nor destroyed in a chemical reaction, what can happen to the energy

Answers

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A 3.50 g sample of an unknown compound containing only C , H , and O combusts in an oxygen‑rich environment. When the products have cooled to 20.0 °C at 1 bar, there are 4.41 L of CO2 and 3.26 mL of H2O . The density of water at 20.0 °C is 0.998 g/mL.

Answers

Explanation:

First, calculate the moles of CO_(2) using ideal gas equation as follows.

                PV = nRT

or,          n = (PV)/(RT)

                = (1 atm * 4.41 ml)/(0.0821 Latm/mol K * 293 K)      (as 1 bar = 1 atm (approx))

                = 0.183 mol

As,   Density = (mass)/(volume)

Hence, mass of water will be as follows.

                Density = (mass)/(volume)

             0.998 g/ml = (mass)/(3.26 ml)    

                 mass = 3.25 g

Similarly, calculate the moles of water as follows.

        No. of moles = \frac{mass}{\text{molar mass}}

                              =  (3.25 g)/(18.02 g/mol)            

                              = 0.180 mol

Moles of hydrogen = 0.180 * 2 = 0.36 mol

Now, mass of carbon will be as follows.

       No. of moles = \frac{mass}{\text{molar mass}}

          0.183 mol =  (mass)/(12 g/mol)            

                              = 2.19 g

Therefore, mass of oxygen will be as follows.

              Mass of O = mass of sample - (mass of C + mass of H)

                                = 3.50 g - (2.19 g + 0.36 g)

                                = 0.95 g

Therefore, moles of oxygen will be as follows.

          No. of moles = \frac{mass}{\text{molar mass}}

                               =  (0.95 g)/(16 g/mol)            

                              = 0.059 mol

Now, diving number of moles of each element of the compound by smallest no. of moles as follows.

                         C              H           O

No. of moles:  0.183        0.36       0.059

On dividing:      3.1           6.1            1

Therefore, empirical formula of the given compound is C_(3)H_(6)O.

Thus, we can conclude that empirical formula of the given compound is C_(3)H_(6)O.