A 2.400-g sample of meat is subjected to Kjeldahl analysis. The liberated NH3(g) is absorbed by adding 50.00 mL of H2SO4(aq), which is more than enough. This occurs through the reaction 2NH3(g)+H2SO4(aq)→(NH4)2SO4(aq) The excess acid requires 19.90 mL of 0.5510 M NaOH for its complete neutralization.

51.04% of protein in the sample

Explanation:

The question is the percentage of protein in the sample and the acid is 0.2496 M  H2SO4.

First we need to calculate the moles of sulfuric acid that react:

Moles of sulfuric acid added:

0.0500L * 0.2496mol/L = 0.01248 moles of sulfuric acid.

Moles in excess:

0.01990L * 0.5510mol/L = 0.01096 moles NaOH * (1 mol H2SO4 / 2 moles NaOH) = 0.00548 moles of sulfuric acid.

Moles that react:

0.01248 moles - 0.00548 moles = 7.0x10⁻³ moles of H2SO4 react.

Based on the reaction, the moles of NH3 = Moles of N are:

7.0x10⁻³ moles of H2SO4 * (2moles NH3 / 1 mole H2SO4) = 0.014 moles NH3 = Moles N

The mass of nitrogen (Molar mass: 14g/mol):

0.014 moles N * 14g/mol = 0.196g of N.

The percentage of N is:

0.196g N / 2.400g = 8.17% of N

6.25 times the percentage of nitrogen is the percentage of protein:

8.17%*6.25 =

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"