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Doss [256]
3 years ago
8

What is the mass of 2.25 moles of manganese(lv) sulfide(MnS2)

Chemistry
2 answers:
sertanlavr [38]3 years ago
8 0
Hi! I'm hoping I understand your question correctly.. To calculate the mass or grams of MnS^2 you will have to find the molar mass. The molar mass is 87.003 g/mol. All you have to do is use stoichiometry and multiply that molar mass by the mols (2.25)! Hope this helped! Remember your sig figs, too!
Alona [7]3 years ago
5 0

Explanation:

Number of moles is defined as mass divided by molar mass of the substance or atom.

Mathematically,     No. of moles = \frac{mass}{molar mass}

It is known that molar mass of MnS_{2} is 119.058 g/mol and number of moles is given as 2.25 mol.

Therefore, calculate mass of MnS_{2} as follows.

                          No. of moles = \frac{mass}{molar mass}

                                   2.25 mol = \frac{mass}{119.058 g/mol}

                                 mass = 267.88 g

Thus, we can conclude that mass of 2.25 moles of manganese(lv) sulfide(MnS2) is 267.88 g.

                 

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sertanlavr [38]

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Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

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where,

q = heat released by the reaction

q_1 = heat absorbed by the calorimeter

q_2 = heat absorbed by the water

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c_2 = specific heat of water = 4.18J/g^oC

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Now put all the given values in the above formula, we get:

q=[(15.8J/^oC\times 8.1^oC)+(100.0g\times 4.18J/g^oC\times 8.1^oC)]

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Now we have to calculate the enthalpy change for the solution.

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Now,

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