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2 years ago

How many moles of Mg3(PO4)2 are in 350.00 grams of Mg3(PO4)2?

Chemistry

2 answers:

dusya [7]2 years ago

6 0

Answer:

1.3 moles/ 1.33150727 moles

Explanation:

350g x 1 mol/262.86g = 1.3 moles

jok3333 [9.3K]2 years ago

6 0

The molar mass of Mg3(PO4)2 is 262.855 g/mol

Atomic weight of the parts:

Mg 24.3050 x 3 = 72.92

P 30.973762 x2 = 61.95

O 15.9994 x8 = 127.995

There are 262.855 grams in one mole, so

350.00 g x (262.855g/1 mole) = 1.33 moles Mg3(PO4)2

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What is the formula for ammonium carbonite? How did you get it?

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Answer: Carbon Dioxide, Methane, Nitrous Oxide and Fuorinated Gases

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3 years ago

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3 years ago

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4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

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1. The volume of ammonia consumed in the reaction is 23.2 L

2. The volume of oxygen consumed in the reaction is 29 L

<h3>Balanced equation</h3>

4NH₃(g) + 5O₂(g) --> 4NO(g) + 6H₂O(g)

From the balanced equation above,

6 L of H₂O were produced from 4 L of NH₃ and 5 L of O₂

<h3>1. How to determine the volume of ammonia, NH₃ consumed</h3>

From the balanced equation above,

6 L of H₂O were produced from 4 L of NH₃

Therefore,

34.8 L of H₂O will be produced from = (34.8 × 4) / 6 = 23.2 L of NH₃

Thus, 23.2 L of NH₃ were consumed

<h3>2. How to determine the volume of oxygen, O₂ consumed</h3>

From the balanced equation above,

6 L of H₂O were produced from 5 L of O₂

Therefore,

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Thus, 29 L of O₂ were consumed

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1 year ago