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trasher [3.6K]
3 years ago
13

A balloon contains 5 L of oxygen at a pressure of 90 kPa. How much oxygen will the balloon contain if the pressure is lowered to

67 kPa?
Chemistry
1 answer:
laila [671]3 years ago
7 0
Answer is: volume will be 6,7 L.
Boyle's Law: the pressure volume law - <span> volume of a given amount of gas held  varies inversely with the applied pressure when the temperature and mass are constant.
p</span>₁V₁ = p₂V₂.
90 kPa · 5 L = 67 kPa · V₂.
V₂ = 90 kPa · 5 L / 67 kPa.
V₂ = 6,7 L, but same amount of oxygen.


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<em>Step 1.</em> Write the chemical equation

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<em>Step 2.</em> Calculate the moles of Fe^(2+)

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How many liters of hydrogen gas will be produced at STP from the reaction of 7.179 x 1023 atoms of magnesium with 54.21 g of pho
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<h3>Answer:</h3>

18.58 liters of hydrogen gas

<h3>Explanation:</h3>

We are given;

  • The equation;

3Mg + 2H₃(PO₄) → Mg₃(PO₄)₂ + 3H₂

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We are required to determine the volume of hydrogen gas produced;

Step 1; moles of Magnesium

1 mole of an element contains 6.02 × 10^23 atoms

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Moles of Mg = (7.179 x 10^23 ) ÷ (6.02 × 10^23)

                   = 1.193 moles

Step 2: Moles of phosphoric acid

moles = Mass ÷ Molar mass

Molar mass of phosphoric acid = 97.994 g/mol

Therefore;

Moles of Phosphoric acid = 54.21 g ÷ 97.994 g/mol

                                           = 0.553 moles

Step 3: Determine the rate limiting reagent

From the mole ratio of Mg to Phosphoric acid (3 : 2);

1.193 moles of magnesium requires 0.795 moles of phosphoric acid while,

0.0553 moles of phosphoric acid requires 0.8295 moles of Mg

Therefore, phosphoric acid is the rate limiting reagent

step 4: Determine the moles of hydrogen produced

From the equation, w moles of phosphoric acid reacts to produce 3 moles of hydrogen;

Therefore; moles of Hydrogen = moles of phosphoric acid × 3/2

                                                   = 0.553 moles × 3/2

                                                   = 0.8295 moles

Step 5: Volume of hydrogen gas

1 mole of a gas occupies a volume of 22.4 liters at STP

Therefore;

Volume of Hydrogen = 0.8295 moles × 22.4 L/mol

                                  = 18.58 Liters

Therefore; 18.58 liters of hydrogen gas  will be produced

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