Use the problem below to answer the question.HCl + NaOH → NaCl + H2O1. If 30g of HCl is reacted with excess NaOH, and 10g of NaCl is produced, what is the theoretical yield of the experiment?2. If 30 g of HCl is reacted with excess NaOH, and 10 g of NaCl is produced, what is the percent yield of the experiment?

1. The theoretical yield of an experiment is the maximum amount of product that can be produced from a given amount of reactant. In this case, we need to find out how much NaCl can theoretically be produced from 30g of HCl.

First, we need to find the molar mass of HCl and NaCl. The molar mass of HCl is approximately 36.5 g/mol and the molar mass of NaCl is approximately 58.5 g/mol.

Next, we calculate the number of moles of HCl in 30g. This is done by dividing the mass of HCl by its molar mass:

30g HCl * (1 mol HCl / 36.5 g HCl) = 0.82 mol HCl

Since the reaction is a 1:1 ratio, we can assume that 0.82 mol of NaCl can be produced. We then convert this back to grams using the molar mass of NaCl:

0.82 mol NaCl * (58.5 g NaCl / 1 mol NaCl) = 47.97 g NaCl

So, the theoretical yield of the experiment is 47.97g of NaCl.

2. The percent yield of an experiment is the ratio of the actual yield to the theoretical yield, multiplied by 100%. In this case, the actual yield is 10g of NaCl and the theoretical yield is 47.97g of NaCl. So, the percent yield is:

(10g NaCl / 47.97g NaCl) * 100% = 20.85%

So, the percent yield of the experiment is 20.85%.