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Ksivusya [100]
3 years ago
14

Help please due today

Chemistry
1 answer:
Olegator [25]3 years ago
8 0

Answer:

It is a becaues i am a doctor and i know about that

Explanation:Just here to help

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Calculate the theoretical yield of alum expected from 0.9875 g of aluminum foil. assume the aluminum is the limiting reactant.
skelet666 [1.2K]

Answer: 17.34 grams of alum will be produced if 0.9875 g of Aluminium foil was used.

Explanation: Reaction to form alum from Aluminium is given as:

2Al(s)+2KOH(aq.)+2H_2O(l)+4H_2SO_4(aq.)\rightarrow 2KAl(SO_4)_2(s).12H_2O(l)+3H_2(g)

We are given Aluminium to be the limiting reactant, so the formation of alum will be dependent on Aluminium because it limits the formation of product.

By stoichiometry,

2 moles of Al is producing 2 moles of Alum

Mass of 2 moles of Aluminium = (2 × 27)g/mol = 54 g/mol

Mass of 2 moles of alum = (2 × 474)g/mol = 948 g/mol

54 g/mol of aluminium will produce 948 g/mol of alum, so

\text{0.9875 grams of aluminium will produce}=\frac{948g/mol}{54g/mol}\times 0.9875g

Amount of Alum produced = 17.34 grams

Theoretical yield of alum = 17.34 grams.

5 0
3 years ago
How many moles of ammonia can be produced when 2.50 moles of hydrogen react?
jeyben [28]

Answer:

5/3 moles

Explanation:

3.moles.of H2 gives 3 moles of NH3

so 2.5 moles gives 5/3 moles.

Note:Apply simple. concept of unitary method.

3 0
3 years ago
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