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Help chemistry question!!!! 10 points

[From: ] [author: ] [Date: 11-09-22] [Hit: ]
06 * (M(ironphosphate) = 357.478)= 1808.......
5.06 moles of iron (II) phosphate, Fe3(PO4)2, are produced in a reaction. What mass of iron (II) phosphate is produced?

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Fe3(PO4)2 mol.mass = (Fe:3 x 56) + (P:2 x 31) + (O:16 x 8) = 168 + 62 + 128 = 358g/mol.

mass / mol.mass = moles.
moles x mol.mass = mass = 5.06mol x 358g/mol = 1812g = 1.8kg of iron (II) phosphate.

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n= m/M --> m=n*M --> mFe3PO42 = 5.06 * (M(ironphosphate) = 357.478)= 1808.838 gram
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