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How many grams of #(NH_4)_2SO_4# are present in .100 moles of #(NH_4)_2SO_4# ?

Answer 1

Avery Adams

Ammonium sulfate has an equivalent weight of #132.14# #g*mol^-1#.

You have a #0.100# #mol# quantity.

#0.100# #cancel(mol)# #xx# #132.14# #g*cancel(mol^-1)# #=# #??# #g#.

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Answer 2

Wyatt Atkins

The mass in grams of 0.100 mol #"(NH"_4)_2"SO"_4"# is 13.2 g.

The molar mass of #"(NH"_4)_2"SO"_4"# is needed, which is #"132.13952 g/mol"#. https://tutor.hix.ai

Multiply the provided mole by the molar mass of ammonium sulfate to get the mass of 0.100 mol.

#0.100 cancel(mol (NH_4)_2SO_4)xx(132.13952 g (NH_4)_2SO_4)/(1 cancel(mol (NH_4)_2SO_4))=13.2 g (NH_4)_2SO_4"# (rounded to three significant figures)

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Answer 3

Naomi Aronson

To find the number of grams of (NH₄)₂SO₄ present in 0.100 moles of (NH₄)₂SO₄, you can use the formula:

mass = moles × molar mass

The molar mass of (NH₄)₂SO₄ is calculated as follows:

2(N) + 8(H) + 1(S) + 4(O) = 14.01 × 2 + 1.008 × 8 + 32.06 + 16 × 4 = 132.14 g/mol

Now, plug in the values:

mass = 0.100 moles × 132.14 g/mol = 13.214 grams

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.