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

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kjuli1766 kjuli1766 · Answer 1 · 2022-12-17T10:24:02+01:00

The moles of metal sulphate formed in the given reaction is 0.0598.

The reaction of metal nitrate M(NO3)2 with sodium sulphate Na2SO4 can be written as

M(NO3)2 + Na2SO4 → 2 NaNO3 + MSO4

1 mole of metal nitrate produces 1 mole of metal sulphate.

let molar mass of metal M be 'a' g/mol.

molar mass of M(NO3)2 = a+ 14 × 2 + 16×6 = 124+a

moles of M(NO3)2 reacted = 9.835 / (124+a)

molar mass of MSO4 = a+ 32 + 16×4 = 96+a

moles of MSO4 formed = 8.159 / (96+a)

Since 1 mole of metal nitrate produces 1 mole of metal sulphate thus moles of M(NO3)2 reacted and moles of MSO4 formed should be equal.

⇒ 9.835 / (124+a) = 8.159 / (96+a)

⇒ 9.835 × (96+a) = 8.159× (124+a)

⇒ 944.16 + 9.835a = 1011.716 + 8.159a

⇒ 9.835a - 8.159a = 1011.716 - 944.16

⇒ 1.676a = 67.556

⇒ a = 40.3g/mol

Molar mass of unknown metal M is 40.3g/mol.

moles of MSO4 formed = 8.159 / (96+a)

putting the value of 'a' we get

moles of MSO4 formed = 8.159 / (96+40.3) = 0.0598

Thus moles of metal sulphate formed in the given reaction is 0.0598.

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