1 g of a mixture of nahco3 annd na2co3 was heated at 300 deg celsius.the volume of co2 obtained at stp are found to be 112ml.calculate the % of na2co3 in the mixture.
please help i am not able to understand!!
To calculate the percentage of Na2CO3 in the mixture, we can use the concept of stoichiometry and the ideal gas law to determine the amount of CO2 produced.
First, we need to understand the chemical reaction that occurs when the mixture of NaHCO3 and Na2CO3 is heated:
2NaHCO3(s) -> Na2CO3(s) + CO2(g) + H2O(g)
From the balanced equation, we can see that for every 2 moles of NaHCO3, we get 1 mole of Na2CO3 and 1 mole of CO2. Since the NaHCO3 and Na2CO3 are in a 1:1 ratio, we can assume that the mass percentage of Na2CO3 in the mixture is the same as the mole percentage.
Now, let's calculate the number of moles of CO2 produced using the ideal gas law.
We know that the volume of CO2 obtained at STP (Standard Temperature and Pressure) is 112 mL. At STP, the values are 22.4 L for the molar volume and 273 K for the temperature.
Using the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(K·mol)), and T is the temperature in Kelvin, we can solve for n:
n = PV / RT
Since the volume is given in mL, we need to convert it to liters:
Volume (V) = 112 mL = 0.112 L
Substituting the given values into the equation:
n = (1 atm * 0.112 L) / (0.0821 L·atm/(K·mol) * 273 K)
n = 0.00502 moles
Now, since 1 mole of Na2CO3 produces 1 mole of CO2, the number of moles of Na2CO3 in the mixture is also 0.00502 moles.
To calculate the percentage of Na2CO3 in the mixture, we need to divide the moles of Na2CO3 by the total moles of the mixture:
% Na2CO3 = (0.00502 moles Na2CO3 / Total moles of mixture) * 100
Since we know that the mass percentage of Na2CO3 in the mixture is the same as the mole percentage, we can rewrite the equation as:
% Na2CO3 = (0.00502 moles Na2CO3 / Total moles of mixture) * 100
Given that the total moles of mixture is 1 (as 1g is the total mass of the mixture), we can calculate the percentage:
% Na2CO3 = (0.00502 moles Na2CO3 / 1 mole of mixture) * 100
% Na2CO3 = 0.502%
Therefore, the percentage of Na2CO3 in the mixture is 0.502%.