A subtance X contains 10 gram of calcium carbonate calculate the number of oxygen atom present in 10 gram of calcium carbonate

☃️ Chemical formulae ➝

How to find?

For solving this question, We need to know how to find moles of solution or any substance if a certain weight is given.

And, Then we need to find the number of molecules which can be found by:

Here, Na = Avagadro Number (6.022 × 10²³)

☀️ As it is a compund, We considered molecules. Then, we can predict the no. of atoms of elements in one molecule of compound and get the answer.

Solution:

Atomic weight of elements:

❍ Molecular weight of

= 40 + 12 + 3 × 16

= 52 + 48

= 100 g/mol

❍ Given weight: 10 g

Then, no. of moles,

⇛ No. of moles = 10 g / 100 g mol‐¹

⇛ No. of moles = 0.1 moles

☄ No. of moles of Calcium carbonate in the given weight = 0.1 moles

Finding no. of molecules,

⇛ No. of molecules of CaCO3 = 0.1 × 6.022 × 10²³

⇛ No. of molecules of CaCO3 = 6.022 × 10²²

Now,

Then, 6.022 × 10²² molecules will have,

= 3 × 6.022 × 10²² atoms of oxygen

= 18.022 × 10²² atoms of oxygen

In more scientific form,

= 1.8022 × 10²³ atoms of oxygen

☄ Hence, Final answer - 1.8022 × 10²³ atoms

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given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g