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An unknown metal nitrate with the formula M(NO3), was dissolved in water and treated with excess aqueous sodium sulfate. The total amount of sulfate salt produced in this reaction had a mass of 8.159 g. The initial mass of the metal nitrate sample was 9.835 g. Assuming that the reaction goes to completion, determine the number of moles of the sulfate salt produced. DO NOT include units in your answer. If you round during your calculation, be sure to keep at least three (3) decimal places Do NOT use scientific notation in your answer. Report your answer to three (3) decimal places

Respuesta :

In this reaction had a mass of 8.159 g. The initial mass of the metal nitrate sample was 9.835 g. Assuming that the reaction goes to completion,  the number of moles of the sulfate salt produced is 1 mole.

The reaction is given as :

M(NO₃)₂   +   Na₂SO₄   ------>    2NaNO₃    +   MSO₄

so, it is clear that , 1 mole of metal nitrate produces 1 mole of metal sulphate.

mass of metal nitrate = 9.835 g

mass of metal sulfate = 8.158 g

moles of metal nitrate = moles of the metal sulfate.

moles = mass / molar mass

Thus, the moles of  MSO₄ = 1 mole.

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