4. DISCUSSION conclusions (not confirmatory observations) are justfied by each o

Question

4. DISCUSSION conclusions (not confirmatory observations) are justfied by each of the following observations. and low pressure. State what In a similar a. The properties of the product are different from those of either Si or C b. The product is homogeneous. c. The mass of the product is greater than that of the Si used. d. If an additional amount of either Si or C is added and the mixture is treated as described above a heterogeneous mixture of the original product and the added element is obtained Show all setups. Be careful with significant figures e. Write the equation for the reaction between copper and sulfur which occurred in your crucible. Use the allotropic form of sulfur, Ss, in your equation. Write a separate equation for the reaction of the excess sulfur f. To observe the effect of a small weighing error, calculate the percentage change in the experimental mole ratio caused by an error in the mass of the final heating of the crucible which is 10 mg larger than the experimental value for your first sample. Assume no error in the mass of copper. g. What percent by mass of Sb2S5 is sulfur? h. Fill in the blanks: (don't forget to show numerical setup) 8.300 moles of NH3 are equivalent to grams. i. 0.030 moles of Zn are equivalent tograms. 500. grams of SO3 are equivalent to moles. . 0.110 grams of N204 are equivalent to __ moles. Find the simplest (empirical) formula for each of the following compounds, given the percentage by i. 72.4% Fe and 27.6% O. mass: 0.5% H, 47.6% Pt, and 51.9% Cl. 28.0% N, 10.7% H, 20.6% P, and 42.6% O. ofM. whatisforms an xido,M203 whichcontains 81.9% ofthe metalbyweight. Findthe at micweg the chemical symbol for the metal? t

Explanation / Answer

(e)

Cu + S8 ---> Cu2S

On the reactant side, copper is one and sulphur is eight. But the product turns out to be Cu2S. To balance the equation, add 16 to Cu on the reactant side and add 8 in front of Cu on the product side to complete the equation.

16Cu + S8 ---> 8Cu2S

(h)

Sb2S5

Total Mass = ( 2 x 121.8 g) + (5 x 32 g) =
= (243.6 + 160) g
= 403.6g

Mass of sulfur = 160 g

% mass of sulfue = (160 / 403.6) x 100
  = 39.64%

(h)
(i) molar mass of NH3 = 17 g/mol
So, 1 mole of NH3 = 17 g
or, 8.3 moles of NH3 = 8.3 x 17 g = 141.1 g

(ii) molar mass of Zn = 65.4 g/mol
So, 1 mole of Zn = 65.4 g
or, 0.03 moles of Zn = 0.03 x 65.4 g = 1.962 g

(iii) molar mass of SO3 = 80 g/mol
So, 80 g of SO3 = 1 mole
or, 500 g of SO3 = (500 / 80) mole = 6.25 mol

(iv) molar mass of N2O4 = 92 g/mol
So, 92 g of N2O4 = 1 mole
0.11 g of N2O4 = (0.11 / 92) mole = 0.0012 mol

(i)

(i) 72.4% Fe and 27.6% O

1. Assume you have 100 grams of this stuff.
You then have:
72.4 g of Fe
27.6 g of O

2. Convert each to moles.
72.4 g of Fe/55.85 = 1.296 mol Fe
27.6 g of O/16 = 1.725 mol O

3. Divide by the least moles number.
1.296 mol Fe/1.296 = 1
1.725 mol O/1.296 = 1.33

4. Write out the formula.
FeO1.33
The 1.33 = 4/3
So, just multiply by the lowest common number to get the 4/3 to be a whole number, which happens to be a 3, so...
the empirical formula is Fe3O4.

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