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3 years ago

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What mass of ammonia, NH3, is necessary to react with 2.1 x 10^24 molecules of oxygen in the following reaction? 4NH3(g) + 7O2(g

) --> 6H2O(g) + 4NO2(g)

1 answer:

liubo4ka [24]3 years ago

4 0

33.94
multiply 17.031*1.993

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Read 2 more answers

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An analytical chemist weighs out 0.093g of an unknown monoprotic acid into a 250mL volumetric flask and dilutes to the mark with

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Explanation:

<u>Step 1:</u> The balanced equation

HA(aq) + NaOH(aq) → NaA(aq) + H2O(l)

It takes 1 mole of NaOH to neutralize 1 mole of the triprotic acid. This is called the reaction stoichiometry.

<u>Step 2:</u> Data given

Mass of the acid = 0.093 grams

volume = 250 mL

titrates with 0.16 M NaOH

adds 6.5 mL NaOH

<u>Step 3: </u>Calculate moles of NaOH

We know the concentration and volume of NaOH needed to neutralize the acid.

By determining the moles of NaOH in that volume in liters (95.9mL=0.0959L), the moles of acid in the original sample can be determined from the reaction stoichiometry.

Moles = Molarity * Volume

Moles = 0.16 M * 0.0065 L

Moles = 0.00104 moles NaOH

<u>Step 4: </u>Calculate moles of the unknown acid:

It takes 1 mole of NaOH to neutralize 1 mole of the triprotic acid. This is called the reaction stoichiometry.

For 0.00104 moles NaOH we have 0.00104 moles of HA

<u>Step 5: </u>Calculate the molar mass of the acid

Molar mass Ha = Mass Ha / moles Ha

Molar mass Ha = 0.093 grams / 0.00104 moles

Molar mass Ha = 89.42 g/mol ≈89 g/mol

The molar mass of the unknown acid is 89 g/mol

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