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marta [7]
3 years ago
6

How many grams of K2O will be produced from 0.50 g of K and 0.10 g of O2?

Chemistry
1 answer:
Rudik [331]3 years ago
7 0

Answer:

0.6g

Explanation:

Given parameters:

Mass of K = 0.5g

Mass of O₂  = 0.10g

Unknown:

Mass of K₂O  = ?

Solution:

To solve this problem, let us write the reaction equation first;

                   4K   +   O₂     →     2K₂O

The reaction above delineates the balanced chemical reaction.

To solve this problem, we need to know the limiting reactant. This reactant is the one that determines the amount and extent of the reaction because it is given in short supply. The other reactant is the one in excess.

Start off by find the number of moles of the reactant;

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

         Molar mas of K  = 39g/mol

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

 Number of moles of K  = \frac{0.5}{39}   = 0.013moles

 Number of moles of O₂    = \frac{0.1}{32}   = 0.031moles

From the balanced reaction;

          4 moles of K reacted with 1 mole of O₂

         0.013 moles of K will react with \frac{0.013}{4}   = 0.0078 moles of O₂

We see that oxygen gas is in excess. We were given 0.031moles of the gas but only require 0.0078moles of oxygen gas.

The limiting reactant is potassium.

    therefore;

              4 moles of K produced 2 moles of K₂O

             0.013 moles of K will produce \frac{0.013 x 2}{4}   = 0.0065‬moles of K₂O

to find the mass of K₂O;

   Mass of K₂O  = number of moles x molar mass

                Molar mass of K₂O  = 2(39) + 16  = 94g/mol

  Mass of K₂O = 0.0065 x 94  = 0.6g

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What are dipoles, and what is the difference between a natural dipole and an induced dipole?
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See explanation below.

Explanation:

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At 25.0°c, a solution has a concentration of 3.179 m and a density of 1.260 g/ml. the density of the solution at 50.0°c is 1.249
oksano4ka [1.4K]

Answer: -

3.151 M

Explanation: -

Let the volume of the solution be 1000 mL.

At 25.0 °C, Density = 1.260 g/ mL

Mass of the solution = Density x volume

= 1.260 g / mL x 1000 mL

= 1260 g

At 25.0 °C, the molarity = 3.179 M

Number of moles present per 1000 mL = 3.179 mol

Strength of the solution in g / mol

= 1260 g / 3.179 mol = 396.35 g / mol (at 25.0 °C)

Now at 50.0 °C

The density is 1.249 g/ mL

Mass of the solution = density x volume = 1.249 g / mL x 1000 mL

= 1249 g.

Number of moles present in 1249 g = Mass of the solution / Strength in g /mol

= \frac{1249 g}{396.35 g/mol}

= 3.151 moles.

So 3.151 moles is present in 1000 mL at 50.0 °C

Molarity at 50.0 °C = 3.151 M

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