Consider the following chemical reaction:
NaOH +HC1 -> NaC1 + H20
When 36.5 g of HC1 are reacted with 40.0 g of NaOH , how many grams of water are produced along with 58.5 g of NaC1?
I have no clue
To find the number of grams of water produced in the chemical reaction, we need to use the concept of stoichiometry.
Stoichiometry is the quantitative relationship between the reactants and products in a chemical reaction. It is based on the balanced chemical equation, which gives the ratio of moles between the reactants and products.
Let's start by calculating the number of moles of HC1 and NaOH. We can use the molar mass of each substance to convert grams to moles.
Molar mass of HC1 = 36.5 g/mol
Molar mass of NaOH = 40.0 g/mol
Number of moles of HC1 = mass of HC1 / molar mass of HC1
Number of moles of HC1 = 36.5 g / 36.5 g/mol
Number of moles of HC1 = 1 mol
Number of moles of NaOH = mass of NaOH / molar mass of NaOH
Number of moles of NaOH = 40.0 g / 40.0 g/mol
Number of moles of NaOH = 1 mol
The balanced chemical equation tells us that 1 mol of HC1 reacts with 1 mol of NaOH to produce 1 mol of H2O and 1 mol of NaC1. Therefore, we can conclude that 1 mol of water is produced.
Now, let's calculate the number of grams of water produced using the molar mass of water.
Molar mass of H2O = 18.0 g/mol
Number of grams of water produced = number of moles of H2O x molar mass of H2O
Number of grams of water produced = 1 mol x 18.0 g/mol
Number of grams of water produced = 18.0 g
Therefore, when 36.5 g of HC1 are reacted with 40.0 g of NaOH, 18.0 g of water is produced, along with 58.5 g of NaC1.