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How many grams of nitrogen are there in 237 g of aluminum nitrate?

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First determine the molar mass of aluminum nitrate: Al(NO3)3Assuming it's anhydrous aluminum nitrate, using the atomic masses of the elements: Al: 26.98 N: 3 14.01 O: 9 16.00 Total: 213.01 g/mol The (NO3)3 means that there are 3 NO3 for each Al, so the total number of N in one mole is 3 and there are 339 OFrom this you can determine the number of moles in 237 g: 237 g / 213.01 g/mol 1.11 mol Then use the atomic mass of nitrogen and the number of moles of nitrogen in the compound (3): 1.11 mol 3 14.01 g/mol 46.7 g of nitrogen You can check your work by calculating the other elements and verifying they add to 237Al: 1.11 mol 26.98 g/mol 29.9 g O: 1.11 mol 9 16.00 159.8 g 159.8+29.9+46.7 236.4 (It's a little off due to rounding error)

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