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natka813 [3]
3 years ago
5

The reaction between potassium chlorate and red phosphorus takes place when you strike a match on a matchbox. if you were to rea

ct 65.1 g of potassium chlorate (kclo3) with excess red phosphorus, what mass of tetraphosphorus decoxide (p4o10) could be produced?
Chemistry
1 answer:
Phantasy [73]3 years ago
6 0

Since the problem states that excess amount of red phosphorus is supplied, therefore this only means that all of potassium chlorate are consumed to form the product tetraphosphorus decoxide.

The reaction is given as:

10KClO3(s) + 3P4(s) → 3P4O10(s) + 10KCl(s) 

 

First, we calculate for the amount of KClO3 in terms of moles. The molar mass of KClO3 is 122.55 g / mol.

moles KClO3 = 65.1 g / (122.55 g / mol)

moles KClO3 = 0.5312 mol

 

Using stoichiometric ratio of the balanced reaction, 10 moles of KClO3 produces 3 moles of P4O10, therefore:

moles P4O10 = 0.5312 mol KClO3 (3 moles of P4O10 / 10 moles of KClO3)

moles P4O10 = 0.16 mol

Converting this to mass by multiplying the molar mass of P4O10 = 283.886 g/mol

mass P4O10 = 0.16 mol * 283.886 g/mol

mass P4O10 = 45.24 g

 

ANSWER: 45.24 g P4O10

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When aqueous solutions of K3PO4 and Ba(NO3)2 are combined, Ba3(PO4)2 precipitates. Calculate the mass, in grams, of the Ba3(PO4)
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Answer:

Mass of Ba₃(PO₄)₂ = 0.0361 g

Explanation:

Given data:

Volume of Ba(NO₃)₂ = 1.2 mL (1.2 × 10⁻³ L )

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Molarity of K₃PO₄ =  0.604 M

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3Ba(NO₃)₂  + 2K₃PO₄  → Ba₃(PO₄)₂  + 6KNO₃

Number of moles of Ba(NO₃)₂ = Molarity × Volume in litter

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Number of moles of Ba(NO₃)₂ = 0.182 × 10⁻³ mol

Number of moles of K₃PO₄ = Molarity × Volume in litter

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Number of moles of K₃PO₄ = 2.537 × 10⁻³ mol

Now we will compare the moles of Ba₃(PO₄)₂ with K₃PO₄ and Ba(NO₃)₂ .

              Ba(NO₃)₂        :         Ba₃(PO₄)₂

                   3                :               1

              0.182 × 10⁻³    :              1/3 ×0.182 × 10⁻³ = 0.060 × 10⁻³ mol

                K₃PO₄           :          Ba₃(PO₄)₂

                   2                 :                1

              2.537 × 10⁻³     :               1/2 ×  2.537 × 10⁻³= 1.269 × 10⁻³ mol

The number of moles of Ba₃(PO₄)₂ produced by  Ba(NO₃)₂  are less it will limiting reactant.

Mass of Ba₃(PO₄)₂ = moles × molar mass

Mass of Ba₃(PO₄)₂ = 0.060 × 10⁻³ mol × 601.93 g/mol

Mass of Ba₃(PO₄)₂ = 36.12 × 10⁻³ g

Mass of Ba₃(PO₄)₂ = 0.0361 g

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