A 1.9450 g sample containing only Na2CO3 (F. M. 105.99) and NaHCO3 (F. M. 84.01) is titrated with 35.31 mL of 0.8724 M HCl. Calculate the percentage sodium carbonate and sodium bicarbonate.

1.0979g of Na2CO3

0.8471g of NaHCO3

Explanation:

NaHCO3 and Na2CO3 reacts with HCl as follows:

NaHCO3 + HCl → NaCl + H2O + CO2

Na2CO3 + 2HCl → 2NaCl + H2O + CO2

Moles of HCl required were:

0.03531L * (0.8724mol / L) = 0.03080 moles of HCl

With this information we can write 2 equations:

1.9450g = X + Y (1)

X is the mass of NaHCO3 and Y is mass of Na2CO3

Also we can write:

0.03080 moles = X/84 + 2Y/106 (2)

Where 84 is molar mass of NaHCO3 and 106 is molar mass of Na2CO3, 2 is due the reaction (2 moles of HCl react per mole of Na2CO3)

Replacing (1) in (2):

0.03080 moles = (1.9450-Y)/84 + 2Y/106

0.03080 = 0.023155 - Y/84 + Y/53

0.007645 = 0.00696316Y

Y = 1.0979g of Na2CO3

And mass of NaHCO3 =

1.9450g - 1.0979g =

0.8471g of NaHCO3

a boiling water reactor does not boil the water. a pressurized water reactor boils the water faster. a pressurized water reactor is heated directly by the core. a pressurized water reactor does not use fission.

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.