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OverLord2011 [107]
4 years ago
7

What is the volume occupied by 4 grams of oxygen at room temperature​

Chemistry
1 answer:
son4ous [18]4 years ago
5 0

Answer:

volume of O2 = 3000cm³

Explanation:

no. of moles= mass of oxygen (g)/molar mass of oxygen(g/mol)

                    = 4/32

                    =0.125

volume of oxygen= no. of moles × 24000 cm³ (or 24dm³)

                             = 0.125 × 24000 cm³  

                             =3000cm³  (or 3dm³)

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27.4 g of Aluminum nitrite and 169.9 g of ammonium chloride react to form aluminum chloride, nitrogen, and water. How many grams
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<u>Answer:</u> The mass of excess reagent (ammonium chloride) remained after the reaction is 62.7 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

  • <u>For aluminium nitrite:</u>

Given mass of aluminium nitrite = 27.4 g

Molar mass of aluminium nitrite = 41 g/mol

Putting values in equation 1, we get:

\text{Moles of aluminium nitrite}=\frac{27.4g}{41g/mol}=0.668mol

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Given mass of ammonium chloride = 169.9 g

Molar mass of ammonium chloride = 53.5 g/mol

Putting values in equation 1, we get:

\text{Moles of ammonium chloride}=\frac{169.9g}{53.5g/mol}=3.176mol

The chemical equation for the reaction of aluminium nitrite and ammonium chloride follows:

Al(NO_2)_3+3NH_4Cl\rightarrow AlCl_3+3N_2+6H_2O

By Stoichiometry of the reaction:

1 mole of aluminium nitrite reacts with 3 moles of ammonium chloride

So, 0.668 moles of aluminium nitrite will react with = \frac{3}{1}\times 0.668=2.004mol of ammonium chloride.

As, given amount of ammonium chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, aluminium nitrite is considered as a limiting reagent because it limits the formation of product.

Excess moles of ammonium chloride = (3.176 - 2.004) mol = 1.172 moles

Calculating the mass of ammonium chloride by using equation 1, we get:

Excess moles of ammonium chloride = 1.172 moles

Molar mass of ammonium chloride = 53.5 g/mol

Putting values in equation 1, we get:

1.172mol=\frac{\text{Mass of ammonium chloride}}{53.5g/mol}\\\\\text{Mass of ammonium chloride}=(1.172mol\times 53.5g/mol)=62.7g

Hence, the mass of excess reagent (ammonium chloride) remained after the reaction is 62.7 grams

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