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What is the correct chemical formula for the hydrated crystal MnNO34 xH2O if analysis shows 511 g of MnNO34 and 334 g of H2O?
Wiki User
∙ 6y ago
Calculate in two steps:
- Mn(NO3)4 : M = 25 + 4*(14 + 3*16) = 303 g/mol --> 511g / 303(g/mol) = 1.686 mol Mn(NO3)4 in total
- H2O M=18 g/mol -> 334g / 18(g/mol) = 18.55mol H2O in the same total amount (845 g) Mn(NO3)4.(H2O)x
x = 18.55 / 1.686 = 11
Control:
In total there is 1.686 (mol Mn(NO3)4.(H2O)11) * [303(g/mol) + 11*18(g/mol)] = 1.686*[303+198] = 844.7 gram Mn(NO3)4.(H2O)11
which,when rounded, is close enough to 845 (= 511+334) for the 3-digit accuracy.
Wiki User
∙ 6y ago
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