Can someone help me with my stoichiometry homework, it is due tonight.

1. 1 mole of NaCl.

2. 1 mole of H₂O.

Explanation:

The balanced equation for the reaction is given below:

HCl + NaOH —> NaCl + H₂O

From the balanced equation above,

1 mole of HCl reacted with 1 mole of NaOH to produce 1 mole of NaCl and 1 mole of H₂O.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above, we can see that 1 mole of HCl reacted completely with 1 mole of NaOH. Therefore, HCl is the limiting reactant and NaOH is the excess reactant since we have 3.6 moles of NaOH.

1. Determination of the amount of NaCl produced from the reaction.

NOTE: The limiting reactant is used to determine the maximum amount produced since all of it is consumed in the reaction.

The limiting reactant is HCl and the amount of NaCl produced can be obtained as follow:

From the balanced equation above, we can see that 1 mole of HCl reacted to 1 mole of NaCl.

Thus, 1 mole of NaCl was produced.

2. Determination of the amount of H₂O produced from the reaction.

From the balanced equation above, it is evident that 1 mole of HCl reacted to produce 1 mole of H₂O.

Thus, 1 mole of H₂O was produced.

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%

the electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.