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3 years ago

Is it necessary to know the exact volume of the unknown acid or base to be titrated?

2 answers:

7 0

Yes, It is important to know the volume of Unknown acid or base to be titrated.

Titration is carried out in order to find out the concentration (i.e. molarity) of unknown acid or base. In this process a standard solution of acid or base is taken and is titrated with known volume of of titrant. At end point (neutralization) the amount of standard titrant utilized is calculated and following formula is employed to calculate the unknown concentration of unknown solution.

M₁V₁/n₁ = M₂V₂/n₂

White raven [17]3 years ago

4 0

Answer:

YES

Explanation:

Titration is an analytical method or technique which is used to determine the concentration of an unknown solution, called analyte; by using a solution of known volume and concentration, called titrant.

In this method, the volume of the titrant used to neutralize the given analyte is noted and then the concentration of the unknown solution or the analyte is calculated by using the following equation:

$C_{titrant}\cdot V_{titrant} = C_{analyte}\cdot V_{analyte}$

Here, $C_{titrant}$ is the known concentration of the titrant

$C_{analyte}$ is the unknown concentration of the analyte

$V_{titrant}$ is the volume of the titrant used during the titration

$V_{analyte}$ is the known volume of the analyte

Therefore, in order to determine the concentration of the unknown solution or the analyte, the exact volume taken should be known.

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If 125 g of CaCO3 is mixed and reacted with 125 g of HCl, which reactant is limiting and how many grams of CO2 can be made?

Trava [24]

Answer:

CaCO3 is the limiting reactant

55 g of CO2 is made

Explanation:

First we must put down the reaction equation;

CaCO3(s) + 2HCl(aq) ---------> CaCl2(s) + H2O(l) + CO2(g)

Number of mole of CaCO3 = 125g/100gmol-1 = 1.25 moles

From the reaction equation;

1 mole of CaCO3 yields 1 mole of CO2

Hence 1.25 moles of CaCO3 yields 1.25 moles of CO2

For HCl;

number of moles of HCl = 125g/36.5 g mol-1 = 3.42 moles

From the reaction equation;

2 moles of HCl yields 1 mole of CO2

3.42 moles of HCl yields 3.42 * 1/2 = 1.71 moles of CO2

Hence CaCO3 is the limiting reactant.

Mass of CO2 produced = 1.25g * 44 gmol-1 = 55 g of CO2

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2 years ago

Consider the reaction 2CO * O2 —> 2 CO2 what is the percent yield of carbon dioxide (MW= 44g/mol) of the reaction of 10g of c

Arturiano [62]

Answer:

Y = 62.5%

Explanation:

Hello there!

In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:

$m_{CO_2}^{theoretical}=10gCO*\frac{1molCO}{28gCO}*\frac{2molCO_2}{2molCO} *\frac{44gCO_2}{1molCO_2}\\\\ m_{CO_2}^{theoretical}=16gCO_2$

Finally, given the actual yield of the CO2-product, we can calculate the percent yield as shown below:

$Y=\frac{10g}{16g} *100\%\\\\Y=62.5\%$

Best regards!

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2 years ago