In an experiment, a 88.11 mL sample of unknown silver nitrate solution was treated with 9.753 g of sodium chloride, resulting in 4.576 g of precipitate. Calculate the molarity of the silver nitrate solution

Explanation:

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In this case, according to the following chemical reaction:

It is possible to compute the moles of silver nitrate via stoichiometry that produced 4.576 g of silver chloride as shown below:

Thus, since the molarity is obtained by dividing moles by volume, we obtain:

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2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

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