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Since we are looking at the amounts of reactants and products, we first need to write a balanced chemical reaction. For an acid-base reaction, we will see salt and water.
HCl + NaOH --> NaCl + H2O
Now that we have the reaction, we need to check that the equation is balanced which it is. Since we are given the amount of two different reactants, we have to see which of these is the limiting reagent which will limit the amount of product that is made.
First we assume that HCl is limiting and see how much NaCl can be made.
22.4 mL (1 L/1000 mL) (2.00 mol/L HCl) (1 mol NaCl/1 mol HCl) (58.5 g/ mol NaCl) = 2.62 g of NaCl
Now, we have to assume that NaOH is limiting and see how much NaCl can be produced
1.74 g NaOH (1 mol NaOH/ 40 g) (1 mol NaCl / 1 mol NaOH) (58.5 g NaCl/mol) = 2.54 g NaCl
Since 2.54 g is the lower of the two masses, this is the amount of NaCl that can be produced in the reaction.
The reaction between sadium chloride and the hydrochloric acid can be illustrated by,
NaOH + HCl --> NaCl + H2O
so here 1 mole of NaOH reacts with 1 mole of HCl. Hence it is required to convert the amount of HCl and NaOH in to moles.
Number of HCl moles
Concentration of the solution = 2.00M = 2 mol/ liter
Volume of the solution = 22.4 ml
Number of HCl moles = (2mol/1000 ml)*22.4 ml
= 0.0448 mol
Number of NaOH moles
The following molecular weights are considered in this calculation
Na - 23 g/mol , Cl - 35.5 g/mol , O - 16 g/mol , H - 1g/mol
Hence the molecular weight of NaOH = 40g/mol (23 +16+1)
Amount of NaOH in the reaction = 1.76 g
Amount of moles of NaOH in the reaction = 1.74 g / (40 g/mol)
= 0.0435 mol
In this reaction NaOH and HCl reacts in 1:1 ratio.That is one mole of NaOh will react with 1 mole of HCl. As the amount of NaOH moles are less than the HCl moles, 0.0435 mol of HCl will react and the excess 0.0013 mol will be left in the solution.
According to the reaction 1 mole of NaOH will produce 1 mole of NaCL.
Therefore 0.044 mol of NaOH will produce 0.0435 mol of NaCl.
molecular weight of NaCl = 58.5 g/mol (23 + 35.5)
weight of the produced NaCl = (58.5 g/mol0 x (0.0435 mol)
= 2.54 g
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