what is the mass of each element are presenf in 7.15 moles Mg(NO³)²?

**Answer:**

Would you agree that if we got a 4.12∙mol4.12•mol quantity of magnesium nitrate, i.e. a 611.0∙g611.0•g mass (and how did I find that…?), then we got 4.12∙mol×24.3∙g∙mol−1=100.1∙g4.12•mol×24.3•g•mol−1=100.1•g with respect to magnesium, 4.12∙mol×2×14.01∙g∙mol−1=115.4∙g4.12•mol×2×14.01•g•mol−1=115.4•g, with respect to nitrogen, and 4.12∙mol×6×16.0∙g∙mol−1=395.5∙g4.12•mol×6×16.0•g•mol−1=395.5•g, with respect to oxygen… And do the individual elemental masses SUM to 611.0∙g611.0•g? Should they do so? And don’t trust my ‘rifmatik….

**Explanation:**

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