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jonny [76]
3 years ago
9

How many grams of water are needed to form 21.8 grams of oxygen gas?

Chemistry
1 answer:
GaryK [48]3 years ago
5 0

Answer:

24.5g

Explanation:

Given parameters:

Mass of oxygen gas = 21.8g

Unknown:

Mass of water  = ?

Solution:

This problem deals with the decomposition of water;

    Reaction equation;

                  2H₂O →  2H₂  +  O₂

2 mole of water will produce 1 mole of oxygen gas

First find the number of mole of the oxygen gas given;

          Number of moles  = \frac{mass}{molar mass}

    molar mass of O₂   = 2(16) = 32g/mol

 

  Number of moles of O₂  = \frac{21.8}{32}   = 0.68moles

 

    1 mole of oxygen gas will be produced from 2 mole of water

     0.68moles of oxygen gas will be produce from  0.68 x 2 moles of water;   = 1.36moles of water

Mass of water = number of moles x molar mass

  Molar mass of water = 2(1) + 16  = 18g/mol

Mass of water  = 1.36 x 18  = 24.5g

           

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Molarity = moles of solute/volume of solution in liters.

The solute here is NaCl, of which we have 46.5 g. To calculate the molarity of an NaCl solution, we need to know the number of moles of NaCl. To convert from grams to moles, we divide the mass by the molar mass of NaCl. The molar mass of NaCl is the sum of the atomic masses of Na and Cl: 23 amu + 35 amu = 58 amu. For our purposes, we can regard amu as equivalent to grams/mole.

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