If 175g of sodium reacts with excess magnesium nitrate then how many miles of sodium nitrate will form?

12.5 moles NaNO3

Explanation: solution attached.

First balance predict the products

Na+ Mg( NO3)2 => Mg + NaNO3

Next balance the chemical equation

2 Na + Mg(NO3)2 => Mg + 2 NaNO3

Solve for the moles of Na

175 g Na x 1 mole Na/ 14 g Na

Next compare the mole ratios of Na and NaNO3 based from the balanced equation.

2 moles Na : 2 moles NaNO3

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%