16.7 grams of chlorine gas are allowed to react with with 3.31 grams of water

What is the maximum amount of hydrochloric acid that can be formed in grams.
What is the maximum amount of hydrochloric acid that can be formed? grams

What is the FORMULA for the limiting reagent?

What amount of the excess reagent remains after the reaction is complete? grams

Work the problem as a stoichiometry problem, first with 16.7 g Cl and next with 3.31 g H2O. The smaller value for the product is the one that will be correct. Do that, get an answer, show your work, and I'll help you through the last part of how much of the excess reagent remains.

55.277

To find the maximum amount of hydrochloric acid that can be formed, we need to determine the limiting reagent. The limiting reagent is the reactant that is completely consumed in the reaction, thus limiting the amount of product that can be formed.

To determine the limiting reagent, we can use the concept of stoichiometry. First, we need to write a balanced chemical equation for the reaction between chlorine gas (Cl2) and water (H2O):

Cl2 + H2O → HCl + HClO

According to the balanced equation, one mole of Cl2 reacts with one mole of H2O to produce two moles of HCl. Therefore, we need to convert the given masses of chlorine gas and water to moles in order to compare their ratios.

The molar mass of Cl2 is 70.906 g/mol, and the molar mass of H2O is 18.015 g/mol.

For chlorine gas:
Number of moles of Cl2 = mass of Cl2 / molar mass of Cl2
Number of moles of Cl2 = 16.7 g / 70.906 g/mol = 0.235 moles

For water:
Number of moles of H2O = mass of H2O / molar mass of H2O
Number of moles of H2O = 3.31 g / 18.015 g/mol = 0.183 moles

Now, we can compare the ratios of moles of Cl2 and H2O. Since the balanced equation states that the ratio is 1:1, we can see that the limiting reagent is the reactant with the smaller number of moles, which in this case is water (H2O).

To find the maximum amount of hydrochloric acid formed, we can use the mol-to-mol ratio between water and HCl from the balanced equation. The balanced equation states that one mole of water produces two moles of HCl.

Number of moles of HCl = 2 * number of moles of H2O
Number of moles of HCl = 2 * 0.183 moles = 0.366 moles

Now, we can convert the moles of HCl to grams by multiplying by the molar mass of HCl. The molar mass of HCl is 36.461 g/mol.

Mass of HCl = number of moles of HCl * molar mass of HCl
Mass of HCl = 0.366 moles * 36.461 g/mol = 13.348 g

Therefore, the maximum amount of hydrochloric acid that can be formed is 13.348 grams.

To find the formula for the limiting reagent, we can look at the balanced equation. In this case, the limiting reagent is water (H2O).

To determine the amount of the excess reagent that remains after the reaction is complete, we need to calculate the amount of the excess reagent that was not used in the reaction. In this case, chlorine gas (Cl2) is the excess reagent.

First, let's calculate the number of moles of Cl2 that reacted:
Number of moles of Cl2 reacted = number of moles of H2O consumed * (1 mole Cl2 / 1 mole H2O)
Number of moles of Cl2 reacted = 0.183 moles * (1 mole Cl2 / 1 mole H2O) = 0.183 moles

Now, let's determine the number of moles of Cl2 that were in excess:
Number of moles of excess Cl2 = total moles of Cl2 - moles of Cl2 reacted
Number of moles of excess Cl2 = 0.235 moles - 0.183 moles = 0.052 moles

Finally, we can calculate the mass of the excess Cl2 that remains by multiplying the number of moles by the molar mass of Cl2:
Mass of excess Cl2 = number of moles of excess Cl2 * molar mass of Cl2
Mass of excess Cl2 = 0.052 moles * 70.906 g/mol = 3.678 g

Therefore, the amount of the excess reagent (chlorine gas) that remains after the reaction is complete is 3.678 grams.