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mylen [45]
2 years ago
11

What would be the mass in grams of 0.300 moles of the ionic compound formed when magnesium metal reacts with oxygen

Chemistry
1 answer:
yarga [219]2 years ago
4 0

Answer:

Balanced Equation for reaction between Magnesium and Oxygen:

2Mg + O₂ --> 2MgO

Molar Mass of MgO is the atomic masses listed on the periodic table for the two elements Magnesium and Oxygen. Magnesium's molar mass is 24g/mol, and Oxygen's molar mass is 16g/mol. So MgO's molar mass would be 24 + 16 = 40g/mol.

The equation to find moles is:

moles \ = \ \frac{mass}{molar \ mass}

So if we rearrange this equation to find for mass:

mass = moles \ * \ molar \ mass

So you have to multiply 0.300 moles by 40, which gives you 12g

Meaning the mass of Magnesium Oxide is 12g.  

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CH4 with pressure 1 atm and volume 10 liter at 27°C is passed into a reactor with 20% excess oxygen, how many moles of oxygen is
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Answer : The moles of O_2 left in the products are 0.16 moles.

Explanation :

First we have to calculate the moles of CH_4.

Using ideal gas equation:

PV=nRT

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Now put all the given values in the ideal gas equation, we get:

(1atm)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)

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The balanced chemical reaction will be:

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From the balanced reaction we conclude that,

As, 1 mole of CH_4 react with 2 moles of O_2

So, 0.406 mole of CH_4 react with 2\times 0.406=0.812 moles of O_2

Now we have to calculate the excess moles of O_2.

O_2 is 20 % excess. That means,

Excess moles of O_2 = \frac{(100 + 20)}{100} × Required moles of O_2

Excess moles of O_2 = 1.2 × Required moles of O_2

Excess moles of O_2 = 1.2 × 0.812 = 0.97 mole

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Moles of O_2 left in the products = Excess moles of O_2 - Required moles of O_2

Moles of O_2 left in the products = 0.97 - 0.812 = 0.16 mole

Therefore, the moles of O_2 left in the products are 0.16 moles.

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