The balanced equation for the reaction is as follows
Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂O
stoichiometry of Na₂CO₃ to HCl is 1:2
number of Na₂CO₃ moles reacted = molarity x volume
number of Na₂CO₃ moles = 0.100 mol/L x 0.750 L = 0.0750 mol
according to molar ratio of 1:2
1 mol of Na₂CO₃ reacts with 2 mol of HCl
then 0.0750 mol of Na₂CO₃ mol reacts with - 2 x 0.0750 = 0.150 mol
molarity of given HCl solution is 1.00 mol/L
molarity is defined as the number of moles of solute in 1 L of solution
there are 1.00 mol in 1 L of solution
therefore there are 0.150 mol in - 0.150 mol / 1.00 mol/L = 0.150 L
volume of HCl required is 0.150 L
The solution 550 ml total and first we will find the amount of alcohol. 3% = 0.03 550 ml x .03 = 16.5 ml alcohol
Then to find the amount of water used, we just have to subtract the amount of alcohol from the total volume
550 ml total - 16.5 ml alcohol = 533.5 ml water
The minimum mass of NaHCO3 that must be added to the spill to neutralize the acid is 27.216 grams
<h3>calculation </h3>
write the balanced chemical equation
2NaHCO3 +H2SO4 → Na2SO4 +2H2O +2CO2
find the moles of H2SO4 = molarity x volume in liters
volume in liters = 27/1000=0.027 l
moles is therefore= 0.027 x6=0.162 moles
by use of mole ratio of NaHCO3: H2SO4 which is 2:1 the moles of NaHCO3=0.162 x2=0.324 moles
mass of NaHCO3= moles of NaHCO3 x molar mass of NaHCO3(84g/mol)
= 84g/mol x 0.324=27.216 grams
Answer:
by measuring the circumference, diameter, and radius of both volleyballs.
