A sample of titrations of 0.200 M sodium hydroxide (NaOH) being added to 30.0 mL of hydrochloric acid (HCl) of unknown concentration. Measure the volume of sodium hydroxide needed to neutralize the HCl, and calculate the concentration of the HCl.

Assuming 20.0 mL of 0.2 M NaOH is used to neutralize 30 mL of the HCl of unknown concentration, the concentration of the hydrochloric acid is 0.133 M.

Note: The volume of NaOH used isnot given and issp to be determined by titration.

Explanation:

The question is a lab activity. Therefore, the volume of 0.2 M NaOH required for the neutralization of 30.0 mL of hydrochloric acid of unknown concentration is to be determined from titration. However, assuming a certain volume of the base is used for the complete neutralization of the acid, the concentration of the acid can be calculated.

Assuming 20.0 mL of 0.2 M NaOH is used to neutralize 30 mL of the HCl of unknown concentration, the concentration of the hydrochloric acid can be calculated thus:

Step 1: Balanced chemical equation of reaction:

HCl + NaOH ---> NaCl + H₂O

From the equation of the reaction, 1 mole of NaOH reacts with 1 mole of HCl to produce 1 mole of NaCl and water.

Step 2: Determine the number of moles of NaOH in 20.0 mL of 0.200 M solution

Molarity = number of moles/volume in Litres

Number of moles = molarity × volume in litres

Volume of NaOH in litres = 20 ml × 1 litre/1000ml = 0.02 litres

Number of moles NaOH = 0.02 L × 0.2 M = 0.004 moles

Therefore, 0.00 M of NaOH reacted with 0.004 moles of HCl

Step 3: Determine concentration of HCl

Molarity or concentration = number of moles / volume in litres

Number of moles of HCl = 0.004 moles

Volume of HCl in litres = 30.0 mL × 1 L/1000 mL = 0.03 L

Concentration of HCl = 0.004 moles / 0.03 L = 0.133 M

Therefore, assuming 20.0 mL of 0.2 M NaOH is used to neutralize 30 mL of the HCl of unknown concentration, the concentration of the hydrochloric acid is 0.133 M.

1. they vary with a higher number because boiling is way hotter than melting

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2