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Lorico [155]
3 years ago
14

Can I please get help with this question ‍♂️?

Chemistry
1 answer:
guapka [62]3 years ago
3 0

Answer:

1.50 moles of POH or Potassium Hydroxide.

Explanation:

First of all, find the substance formula

The substance formula of Potassium Hydroxide is

KOH

Second of all, if you want to convert from grams to moles, you use molar mass of KOH

Third of all, find the molar mass of KOH.

K = 39.1 amu

O = 16.0 amu

H= 1.0 amu

Potassium has 1 atom, oxygen has 1 atom, and hydrogen has 1 atom.

So do this:

39.1(1) + 16.0(1) + 1.0(1) = 56.1

The molar mass of KOH is 56.1 g/mol.

Fourth of all, use dimensional analysis to show your work.

84.20 grams of KOH * 1 mol/56.1 g/mol

The moles will cancel out.

84.20 divided by 56.1 = 1.500892166

But I have to round my answer to two digits after the decimal points instead of using sig figs.

So I round to the hundredths place

1.500892166 = 1.50

So the final answer is 1.50 moles(don't forget the units) of POH.

Hope it helped!

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Ammonia and oxygen react to form nitrogen and water.
Nata [24]

Answer:

A. 19.2 g of O2.

B. 3.79 g of N2.

C. 54 g of H2O.

Explanation:

The balanced equation for the reaction is given below:

4NH3(g) + 3O2(g) → 2N2+ 6H2O(g)

Next, we shall determine the masses of NH3 and O2 that reacted and the masses of N2 and H2O produced from the balanced equation.

This is illustrated below:

Molar mass of NH3 = 14 + (3x1) = 17 g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68 g

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 3 x 32 = 96 g

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 from the balanced equation = 2 x 28 = 56 g

Molar mass of H2O = (2x1) + 16 = 18 g/mol

Mass of H2O from the balanced equation = 6 x 18 = 108 g

Summary:

From the balanced equation above,

68 g of NH3 reacted with 96 g of O2 to produce 56 g of N2 and 108 g of H2O.

A. Determination of the mass of O2 needed to react with 13.6 g of NH3.

This is illustrated below:

From the balanced equation above,

68 g of NH3 reacted with 96 g of O2.

Therefore, 13.6 g of NH3 will react with = (13.6 x 96)/68 = 19.2 g of O2.

Therefore, 19.2 g of O2 are needed for the reaction.

B. Determination of the mass of N2 produced when 6.50 g of O2 react.

This is illustrated below:

From the balanced equation above,

96 g of O2 reacted to produce 56 g of N2.

Therefore, 6.5 g of O2 will react to produce = (6.5 x 56)/96 = 3.79 g of N2.

Therefore, 3.79 g of N2 were produced from the reaction.

C. Determination of the mass of H2O formed from the reaction of 34 g of NH3.

This is illustrated below:

From the balanced equation above,

68 g of NH3 reacted to 108 g of H2O.

Therefore, 34 g of NH3 will react to produce = (34 x 108)/68 = 54 g of H2O.

Therefore, 54 g of H2O were obtained from the reaction.

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