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larisa86 [58]
2 years ago
6

To completely convert 9. 0 moles of hydrogen gas (h2) to ammonia gas, how many moles of nitrogen gas (n2) are required?

Chemistry
1 answer:
evablogger [386]2 years ago
3 0

To completely convert 9. 0 moles of hydrogen gas (h2) to ammonia gas, 3.0 moles of nitrogen gas (n2) are required.

<h3>What are moles?</h3>

The mole is a SI unit of measurement that is used to calculate the quantity of any substance.

<h3 />

The given reaction is \rm  N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)

By the stoichiometry rule of ratio hydrogen: nitrogen

3 : 1

The reacted moles of nitrogen is equals to H/3 moles of reacted hydrogen

So, moles of nitrogen  

\rm Moles\; of\; nitrogen = \dfrac{9.0 }{3} =3.0\;mol

Thus, 3.0 moles of nitrogen gas (n2) are required.

Learn more about moles

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A and B are powders, A is insoluble while B dissolves to give a pH 3 solution. Mixing A and B gives bubbles or effervescence and
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20 What is the base-conjugate acid pair when a strong acid is added to methylamine, CH3NH2?
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base CH3NH2; conjugate  acid: CH3NH3 +

                 Explanation

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6 0
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In a student experiment, the empirical formula of a copper halide was found by adding aluminum metal to an aqueous solution of t
irina [24]
<h3>Answer: </h3>

The Empirical formula of copper chloride based on the experiment data is CuCl₃

<h3>Solution and Explanation: </h3>
  • Empirical formula of a compound is the formula that shows simplest whole number ratio of the atoms contained in a particular compound.
  • It may be found when the percentage composition by mass or the mass of each atom in the compound is known.

In this case we first determine the mass of each atom in the compound from data given.

<h3>Step 1: Average mass of copper. </h3>

Mass of copper in Trial A = 1.694 - 0.908 = 0.786  

Mass of copper in Trial B = 1.693 - 0.922 = 0.771  

Mass of copper in Trial C = 1.588 - 0.919 = 0.669  

Average mass of copper = (0.786 + 0.771 + 0.669) ÷ 3  

                                         = 0.742 g of Copper

<h3>Step 2: Average Volume of copper chloride  </h3>

Volume in Trial A = 49.6 ml

Volume in Trial B = 48.3 ml

Volume in Trial C = 42.2 ml

Average volume of copper chloride = (49.6 + 48.3 + 42.2) ÷ 3

                                                            = 46.7 ml or  

                                                            = 0.046 L

But we are given 42.62 g of copper chloride in a liter, therefore;

<h3>Step 3: Mass of copper Chloride in the average Volume  </h3>

Using proportions  

1 L = 42.62 g Copper chloride  

0.046 L will contain;  

= (42.62 g CuClₓ × 0.0467 L) ÷ 1

= 1.99 g Copper Chloride  

<h3>Step 4: Mass of chlorine </h3>

Mass of chlorine = Mass of copper chloride – mass of copper  

                           = 1.99 g – 0.742 g  

                          = 1.24 g chlorine  

<h3>Step 5: Moles of each atom in the compound (copper chloride) </h3>

Moles = mass / atomic mass  

Moles of copper = 0.742 g / 63.55 g/mol  

                          = 0.0117 moles copper  

Moles of Chlorine = 1.24 g/35.45 g/mol  

                             = 0.0350 moles Chlorine  

<h3>Step 6: Mole ratio  </h3>

Copper: Chlorine

0.0117 moles : 0.0350 moles  

     = 1 : 2.98  

    = 1 : 3 (whole number ratio)

Therefore, the empirical formula of copper chloride is CuCl₃

 

7 0
3 years ago
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