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Schach [20]
3 years ago
8

A student preforms a chemical reaction in which 35 grams of hydrogen and 65 grams of oxygen reacted to form water. What is the m

ass of the products?
Chemistry
1 answer:
tatuchka [14]3 years ago
8 0

Answer:

Mass of water = 73.08 g

Explanation:

Given data:

Mass of hydrogen = 35 g

Mass of oxygen = 65 g

Mass of water = ?

Solution:

First of all we will write the balanced chemical equation:

2H₂  + O₂   →    2H₂O

Number of moles of hydrogen = mass/ molar mass

Number of moles of hydrogen =  35 g/ 2 g/mol

Number of moles of hydrogen = 17.5 mol

Number of moles of oxygen = 65 g / 32 g/mol

Number of moles of oxygen = 2.03 moles

Now we compare the moles of water with moles hydrogen and oxygen.

                                     H₂               :              H₂O

                                      2                :                2

                                    17.5              :              17.5

                                       O₂             :            H₂O

                                      1                :               2

                                      2.03         :             2× 2.03 =4.06 mol

Number of moles of water produced by oxygen are less so oxygen is limitting reactant.

Mass of water:

           Mass of water = number of moles × molar mass

          Mass of water = 4.06 mol × 18 g/mol

           Mass of water = 73.08 g

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A bike rider approaches a hill with a speed of 8.5 m/s. The total mass of the bike and the rider is 85 kg. Find the kinetic ener
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KE=3070.625 J

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<h3>Further explanation</h3>

mass of bike+rider=85 kg

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The equilibrium constant, Kc, for the following reaction is 4.76×10-4 at 431 K. PCl5(g) PCl3(g) + Cl2(g) When a sufficiently lar
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Answer:

Explanation:

Step 1: Data given

The equilibrium constant, Kc, for the following reaction is 4.76 * 10^-4 at 431 K

The equilibrium concentration of Cl2(g) is  0.233 M

Step 2: The balanced equation

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Step 3: The initial concentration

[PCl5]= Y M

[PCl] = 0M

[Cl2] = 0M

Step 4: Calculate the concentration at equilibrium

[PCl5] = Y + X M = Y - 0.233 M

[PCl]= XM = 0.233 M

[Cl2]= XM = 0.233 M

Step 5: Define Kc

Kc =  [Cl2]* [PCl3] / [PCl5]

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The concentration of PCl5 at the equilibrium is 114.05 M

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