How many kilograms of a fertilizer are made of pure p2o5 would be required to supply 1.69 kilogram of phosphorus to the soil?

Decomposition is a reaction in which a compound is broken down into its components. The mass of pure P2O5 is obtained by mass- mole relationship.

The equation of the decomposition of P2O5  is;

Number of moles phosphorus supplied to the soil = = 13.6 moles

From the reaction equation, we can see that;

2 moles  P2O5 yields 1 mole of phosphorus

x moles of  P2O5 yields 13.6 moles of phosphorus

x = 2 mole * 13.6 moles/ 1 mole

x = 27.2 moles

Mass of  pure P2O5 = 27.2 moles * 142 g/moles = 3862 g or 3.862 Kg

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3.89 kg P2O5 must be used to supply 1.69 kg Phosphorus to the soil.

Explanation:

The molecular mass of P2O5 is

P2 = 2* 31 =           62

O5 = 5 * 16 =         80

Molecular Mass = 142

Set up a Proportion

142 grams P2O5 supplies 62 grams of phosphorus

x    kg P2O5        supplies 1.69 kg of phosphorus

Though this might be a bit anti intuitive, you don't have to convert the units for this question. The ratio is all that is important.

142/x = 62/1.69            Cross multiply

142 * 1.69 = 62x           combine the left

239.98 = 62x               Divide by 62

239.98/62 = x

3.89 kg of P2O5 must be used.

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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