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.

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

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