Question in the description, can you please explain how you got the answer, thank you

Question

What is the greatest amount of sodium chloride in moles that can be made with 13 grams copper II chloride and 15 grams sodium nitrate? Which reactant is limiting? How many moles of the excess reactant will be left after the reaction?

CuCl₂  + 2 NaNO₃  Cu(NO₃)₂ + 2 NaCl

Answer 1

Calculate the molecular masses of cooper II chloride and sodium nitrate.

Copper II chloride = 134.45 g/mol

Sodium Nitrate = 84.9947 g/mol

Divide the masses of copper II choride and sodium nitrate by their relative molecular masses to determine how many moles of each of the reactant are available for the reaction.

13g / 134.45 g/mol Copper II chloride= 0.09669 moles Copper II chloride

15g/ 84.9947 g/mol Sodium Nitrate = .17648 moles sodium nitrate

Based on the balanced equation, there is 1 mole copper II chloride used per 2 moles sodium nitrate used. If you divide the number of moles of sodium nitrate by the moles of copper II chloride, it's less than the ratio of two that are necessary to use all of the sodium nitrate. Therefore, copper II chloride is the limiting reactant.

Based on the chemical reaction, there are 2 moles of sodium chloride made per copper II chloride.

So, multiply the number of moles Copper II chloride by two to get the number of moles of sodium chloride.

0.09669 x2 = moles of sodium chloride.

To determine the amount of excess reactant of the sodium nitrate, it is important to understand that ll of the copper to chloride will be used and there is the 1 to 2 mole relationship of sodium nitrate. Subtract the number of moles copper II chloride used from the total moles of available sodium nitrate.

.17648 moles sodium nitrate - 0.09669 moles Copper II chloride = moles of excess sodium nitrate

To determine the mass of the excess reactant, multiply the number of moles of excess sodium nitrate by the molar mass of sodium nitrate

moles of excess sodium nitrate * 84.9947 g/mol