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Zolol [24]
3 years ago
10

If you burn 55.6 g of hydrogen and produce 497 g of water, how much oxygen reacted?

Chemistry
2 answers:
tester [92]3 years ago
6 0

Answer:

441.28 g Oxygen

Explanation:

  • The combustion of hydrogen gives water as the product.
  • The equation for the reaction is;

2H₂(g) + O₂(g) → 2H₂O(l)

Mass of hydrogen = 55.6 g

Number of moles of hydrogen

Moles = Mass/Molar mass

          = 55.6 g ÷ 2.016 g/mol

          = 27.8 moles

The mole ratio of Hydrogen to Oxygen is 2:1

Therefore;

Number of moles of oxygen = 27.5794 moles ÷ 2

                                               = 13.790 moles

Mass of oxygen gas will therefore be;

Mass = Number of moles × Molar mass

Molar mass of oxygen gas is 32 g/mol

Mass = 13.790 moles × 32 g/mol

<h3>          = 441.28 g</h3><h3>Alternatively:</h3>

Mass of hydrogen + mass of oxygen = Mass of water

Therefore;

Mass of oxygen = Mass of water - mass of hydrogen

                          = 497 g - 55.6 g

<h3>                           = 441.4 g </h3>
Lubov Fominskaja [6]3 years ago
5 0
2H2 + O2 → 2H2O
2 mol H2 + 1mol O2 will produce 2 mol H2O
Molar mass H2O = 18.0153g/mol
497g H2O = 23.813 mol H2O produces
This will require 27.587/2 = 13.793 mol O2

Molar mass O2 = 32g/mol
13.793mol = 13.793*32 = 441.37g O2 required.

The easiest way is subtracting 55.6 from 497 which gives you 441.4.
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What volume is occupied by 18.4 g oxygen at 28.0°C and a pressure of 0.998 torr? (round to sig figs)
Alexxandr [17]

Answer:

11.0 dm³

Explanation:

From the question,

Applying

PV= nRT............... Equation 1

Where P = pressure of oxygen gas, V = volume of oxygen gas, n = number of moles of oxygen, R = molar constant, T = Temperature.

make V the subeject of the equation

V = nRT/P............. Equation 2

But,

Number of mole (n) = Mass of oxygen(m)/Molar mass of oxygen(m')

n = m/m'....................... Equation 3

Substitute equation 3 into equation 2

V = mRT/Pm'............. Equation 4

Given: T = 28°C = (28+273) = 301 K, P = 0.998 torr = (0.998×0.00131579) = 1.3132 atm, m = 18.4 g

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Substitute these values into equation 4

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Explanation:

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[Ar]3d^{7}4s^2

For, Co^{3+}, 3 electrons are lost, thus the configuration is:-

[Ar]3d^{6}

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