You are given an unknown mixture containing NaCl and NaHCO3. When you carry out the heating exactly as described in part A, only NaHCO3 decomposes. From the loss in mass, you can calculate the amount of NaHCO3 present in the mixture and its mass percent composition. Here is your data. Incorporate it in your written procedure.

Experimental data Unknown (NaCl + NaHCO3)
Mass of crucible + cover 72.428 g
Mass of crucible + cover + mixture 74.80 g
Mass of crucible + cover + residue (1st heating) 74.25 g
Mass of crucible + cover + residue (2nd heating) 74.24 g
Mass of crucible + cover + residue (3rd heating) 74.24

From the data above, carry out your calculation for finding the mass of mixture, and mass of final solid remaining in the crucible. Using the knowledge that only loss of mass due to CO2 and H2O can be linked to the original mass of NaHCO3, calculate the mass of NaHCO3 and percent of NaHCO3 in the mixture.

1.52g NaHCO3 were in the original mixture.

Mass percent: 64.1%

Explanation:

When NaHCO3 heats it descomposition occurs as follows:

2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g).

The loss in mass is because of the evaporization of CO2 and H2O. As both are in the same porportion, its molar mass is the sum of both compounds (44g/mol + 18g/mol = 62g/mol)

Loss in mass: 74.80g - 74.24g = 0.56g.

In moles:

0.56g * (1mol / 62g) = 0.00903 moles of gas.

As 1 mole of the gases comes from 2 moles of NaHCO3:

Moles NaHCO3:

0.00903 moles of gas * (2 moles NaHCO3 / 1 mole gas) = 0.018 moles NaHCO3.

In grams (Molar mass NaHCO3: 84g/mol):

0.018 moles NaHCO3 * (84g / mol) = 1.52g NaHCO3 were in the original mixture.

The mass of the mixture was:

74.80g - 72.428g = 2.372g

That means mass percent of NaHCO3 is:

(1.52g /  2.372g) * 100 =  64.1%

the dependant variable is spilled water

the independent is the heat

sorry i was late, but i don't know what assignment your talking about, i'm in flvs as well, im in earth space science. but u are behind pace bc every class should be in the 3's or 4'