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Oh boy. That's a toughie.
Group II metal nitrate upon heating will produce a metaloxide and NO2 in 1:1
X(NO3)2 -> XO + NO2
Umm. I'd probably tackle it by finding the mols of the metal nitrate and metal oxide in terms of x.
5/(x+(2(14+16*3)=5-3.29/(x+16)
5/x+124 = 1.71/x+16
5x+80 = 1.71x + 212.04
3.29x = 132.04
x = 40.1
so, Calcium.
The equation for the Group II nitrate thermal decomposotion is as follows: 2X(NO3)2 ----> 2XO + 4NO2 + O2
5.00g 1.71g (all 3.29g)
You could do it using a bit of algebra or through trial and error. So:
if X = Mg, number of moles of Mg(NO3)2 = 5/148.3 = 0.0337 mol.
Therefore mass os MgO produced = 0.0337 x 40.3 = 1.36g, so can´t be Mg.
if X=Ca, number of moles of Ca(NO3)2 = 5/164.1 = 0.0305 mol.
Therefore mass of CaO produced = 0.0305 x 56.1 = 1.71g
Best way will be fastest way! ^^