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astraxan [27]
3 years ago
10

2.93 mol of MgF2 to grams

Chemistry
1 answer:
mojhsa [17]3 years ago
8 0

The mass of 1.72 mol of magnesium fluoride is 107 grams.

To determine the mass of 1.72 mol of magnesium fluoride, we first need the chemical formula of magnesium fluoride. Magnesium forms a +2 ion (Mg+2) and fluoride forms a -1 ion (F-1). Since all compounds formed from ions have to be electrically neutral, we need 2 fluoride ions and 1 magnesium ion. Therefore, the formula for magnesium fluoride is MgF2.

Now we need to determine the molar mass of the compound from the molar mass values from the periodic table. Let's use a table to calculate this molar mass.

Molar mass of MgF2

Element Molar Mass (g/mol) Quantity Total (g/mol)

Mg 24.31 1 24.31

F 19.00 2 38.00

Total molar mass of MgF2 = 24.31 g/mol + 38.00 g/mol = 62.31 g/mol

This is the mass of one mole of the substance. If we have 1.72 mols of it, we multiply 1.72 by 62.31.

1.72 mol (62.31 g/mol) = 107 grams

We rounded to 107 to keep the correct number of significant digits in our answer.

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<em>Full question: Assume no volume change.  If you formed 0.0910 atm of gas, what is the percent yield?</em>

<em />

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Ni + 3HBr → NiBr₃ + 3/2H₂(g)

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For a complete reaction of the 0.084 moles of HBr you need:

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As there are just 0.01738 moles of Ni, the Ni is limiting reactant. Assuming a theoretical yield, moles of H₂ produced are:

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<em>Where P is pressure (0.0910atm)</em>

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<em>R is gas constant (0.082atmL/molK)</em>

<em>T is absolute temperature in Kelvin (30°C + 273.15 = 303.15K)</em>

<em>And n are moles</em>

PV/RT = n

0.0910atm*2.50L/0.082atmL/molK*303.15K = n

0.00915 moles = n

<em />

And percent yield (Produced moles / Theoretical moles * 100) is:

0.00915 moles / 0.02607moles =

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