An alkaline-seltzer tablet contains exactly 100 mg of citric acid, h3c6h5o7, plus some sodium bicarbonate. what mass of sodium bicarbonate is required to consume 100. mg of citric acid by the following reaction? . h3c6h5o7(aq) + 3 nahco3(aq)= 3 h2o(l) + 3 co2(g) + na3c6h5o7(aq).

mass of NaHCO₃  = 0.126 g

Explanation:

balanced chemical equation  

H₃C₆H₅O₇(aq) + 3 NaHCO₃(aq) → 3 H₂O(l) + 3 CO₂(g) + Na₃C₆H₅O₇(aq).

given data

moles of citric acid   = 1 mol  

mas of citric acid = 100 mg

mas of citric acid = 0.1 g

moles of sodium bicarbonate = 3 mol

mass of sodium bicarbonate = ?

Solution

1st we will find out the mole ratio of H₃C₆H₅O₇ and NaHCO₃ from balanced chemical equation

H₃C₆H₅O₇ : NaHCO₃

               1 : 3

Now we find out number of moles of citric acid  

                  moles = mass / molar mass

   moles = 0.1 g / 192.124 g/mol

   moles = 0.0005 mol

now we find out moles (x) of sodium bicarbonate needed for 0.0005 mol of citric acid    

from the balanced chemical equation the mole ratios are:

                  1 : 3

0.0005 mol  :  x

Cross multiply these ratios

    3 × 0.0005 = 1x

          0.0015  = 1 x

          x = 0.0015 mol

Now we will find out the mass of sodium bicarbonate

                  mass   = moles × molar mass

      mass of NaHCO₃  = 0.0015 mol × 84.007 g/mol

      mass of NaHCO₃  = 0.126 g

give me abc and i will get it for u.

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol