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3 years ago

In an experiment between 26.4 grams of Mg and O,, Mgo is produced. The percent

Chemistry

1 answer:

andrew11 [14]3 years ago

7 0

Answer:

39.6 g

Explanation:

The equation of the reaction is;

2Mg(s) + O2(g) --------> 2MgO(s)

To obtain the limiting reactant;

Number of moles in 26.4 g of Mg = 26.4g/24 g/mol = 1.1 moles

If 2 moles of Mg yields 2 moles of MgO

1.1 moles of Mg yields 1.1 * 2/2 = 1.1 moles of MgO

Number of moles in 26.4 g of O2 = 26.4 g/32g/mol = 0.825 moles

If 1 mole of O2 yields 2 moles of MgO

0.825 moles of O2 yields 0.825 moles * 2/1 = 1.65 moles of MgO

Hence Mg is the limiting reactant.

Theoretical yield of MgO = 1.1 moles of MgO * 40 g/mol = 44 g

Percent yield = 90%

Percent yield = actual yield/theoretical yield * 100

Actual yield = Percent yield * theoretical yield/100

Actual yield = 90 * 44/100

Actual yield = 39.6 g

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3 years ago

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2 years ago

How many liters of water can be made from 55 grams of oxygen gas and an excess of hydrogen at a pressure of 12.4 atm and a tempe

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First, we need the no.of moles of O2 = mass/molar mass of O2
= 55 g / 32 g/mol
= 1.72 mol
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2H2 (g) + O2(g) → 2H2O(g)
we can see that the molar ratio between O2: H2O = 1: 2
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