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3 years ago

What is minimum volume of oxygen required to react with 42.5 g of aluminum in the synthesis of aluminum oxide at stp?

1 answer:

6 0

26.833 liters

Aluminum oxide has a formula of Al₂O₃, which means for every mole of aluminum used, 1.5 moles of oxygen is required (3/2 = 1.5).

Given 42.5 g of aluminum divided by its atomic mass (26.9815385) gives 1.575 moles of aluminum.

Since it takes 1.5 moles of oxygen per mole of aluminum to make aluminum oxide, you'll need 2.363 moles of oxygen atoms.

Each molecule of oxygen gas has 2 oxygen atoms, so the moles of oxygen gas will be 2.363/2 = 1.1815

Finally, you need to calculate the volume of 1.1815 moles of oxygen gas.
1 mole of gas at STP occupies 22.7 liters of volume. Therefore,

1.1815 * 22.7 = 26.8 liters of oxygen gas.

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3 years ago

You dissolve 8.65 grams of lead(l) nitrate in water and then you add 2 50 grams of aluminum. This reaction occurs 2AI(S)+ 3Pb(NO

olga55 [171]

Answer: The theoretical yield of solid lead comes out to be 5.408 grams.

Explanation:

To calculate the moles, we use the following equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of Lead nitrate:

Given mass of lead nitrate = 8.65 grams

Molar mass of lead nitrate = 331.2 g/mol

Putting values in above equation, we get:

$\text{Number of moles}=\frac{8.65g}{331.2g/mol}=0.0261moles$

Moles of Aluminium:

Given mass of aluminium = 2.5 grams

Molar mass of aluminium = 27 g/mol

Putting values in above equation, we get:

$\text{Number of moles}=\frac{2.5g}{27g/mol}=0.0925moles$

For the given chemical reaction, the equation follows:

$2AI(s)+3Pb(NO_3)_2(aq.)\rightarrow 3Pb(s)+2AI(NO3)_3(aq.$

By Stoichiometry:

3 moles of lead nitrate reacts with 2 moles of aluminium

So, 0.0261 moles of lead nitrate are produced by = $\frac{2}{3}\times 0.0261=0.0174moles$ of aluminium.

As, the required amount of aluminium is less than the given amount. Hence, it is considered as the excess reagent.

Lead nitrate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of lead nitrate are produces 3 moles of lead metal.

So, 0.0261 moles of lead nitrate will produce = $\frac{3}{3}\times 0.0261=0.0261moles$ of lead metal.

Now, to calculate the grams or theoretical yield of lead metal, we put in the mole's equation, we get:

Molar mass of lead = 207.2 g/mol

Putting values in above equation, we get:

$0.0261mol=\frac{\text{Given mass}}{207.2g/mol}$

Mass of lead = 5.408 grams

Hence, the theoretical yield of solid lead comes out to be 5.408 grams.

8 0

3 years ago

4.5cm wide by 5.750cm long by 1.50cm tall what is the volume of the box

Anni [7]

Answer:

volume of box is 38.81 cm³.

Explanation:

Given data:

Width of box = 4.5 cm

Height of box = 5.750 cm

Length of box = 1.50 cm

Solution:

Formula:

Volume = length × height × width

by putting values,

V = 1.50 cm × 5.750 cm× 4.5 cm

V = 38.81 cm³

Thus, the volume of box is 38.81 cm³.

5 0

2 years ago