Answer:
62.98 % of the sample of hydrate is water
Explanation:
Step 1: Data given
Mass of the sample of a hydrate of sodium carbonate (Na2CO3) = 2.026 grams
After heating, the mass of the sample is 0.750 g
Molar mass H2O = 18.02 g/mol
Step 2: Calculate mass of water
Mass water = mass of hydrate - mass of sample after heating
Mass water = 2.026 grams - 0.750 grams
Mass water = 1.276 grams
Step 3: Calculate mass % percent of water
Mass % of water = (mass of water / total mass hydrate) * 100 %
Mass % of water = (1.276 grams / 2.026 grams) *100 %
Mass % of water = 62.98 %
62.98 % of the sample of hydrate is water
To determine which order of the reaction it is, first we need to calculate the rate of change of moles.
the data is as follows
time 0 40 80 120 160
moles 0.100 0.067 0.045 0.030 0.020
Q1)
for the first 40 s change of moles ;
= -d[A] / t
= - (0.067-0.100)/40s
= 8.25 x 10⁻⁴ mol/s
for the next 40 s
= -(0.045-0.067)/40
= 5.5 x 10⁻⁴ mol/s
the 40 s after that
= -(0.030-0.045)/40 s
= 3.75 x 10⁻⁴ mol/s
k - rate constant
and A is the only reactant that affects the rate of the reaction
rate = k [A]ᵇ
8.25 × 10⁻⁴ mol/s = k [0.100 mol]ᵇ ----1
5.5 x 10⁻⁴ mol/s = k [0.067 mol]ᵇ -----2
divide the 2nd equation by the 1st equation
1.5 = [1.49]ᵇ
b is almost equal to 1
Therefore this is a first order reaction
Q2)
to find out the rate constant(k), we have to first state the equation for a first order reaction.
rate = k[A]ᵇ
As A is the only reactant thats considered for the rate equation.
Since this is a first order reaction,
b = 1
therefore the reaction is
rate = k[A]
substituting the values,
8.25 x 10⁻⁴ mol/s = k [0.100 mol]
k = 8.25 x 10⁻⁴ mol/s /0.100mol
= 8.25 x 10⁻³ s⁻¹
