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