How many mol of carbon atom are in 1.5 mol of alcohol, CH_3CH_2OH? 2. The value

Question

How many mol of carbon atom are in 1.5 mol of alcohol, CH_3CH_2OH? 2. The value of DeltaHdegree for the reaction below is -482 kJ. Calculate the heat (kJ) released to the surroundings when 38.5 g of O_2 (g) reacts with excess CO. 2CO(g) + O_2(g) rightarrow 2CO_2(g) 3. Calculate the number of grams of solute in 500.0 mL of 0.20 M potassium hydroxide. 4. Express/write the weights in grams of each of the following quantities. 4.1 1.5 mol H_2O 4.2 0.5 mol O_2 4.3 0.2 and C_6H_12O_6 4.4 Onc billion Gold (Au) atoms

Explanation / Answer

1)

CH3CH2OH can be written as C2H5OH

we can see that

moles of C = 2 x moles of C2H5OH

moles of C = 2 x 1.5

moles of C = 3

so

there are 3 mol of carbon atoms in 1.5 mol of alcohol

2)

we know that

moles = mass / molar mass

so

moles of 02 = 38.5 / 32 = 1.203125

now

consider the given reaction

2 CO + 02 --> 2 C02

we can see that

1 mole of 02 ---> -482 kJ

1.203125 mole of 02 ---> y kJ

y = -482 x 1.203125

y = -579.9 kJ

so

579.9 kJ of heat is released

3)

we know that

moles = molarity x volume (ml) / 1000

so

moles of KOH = 0.2 x 500 / 1000

moles of KOH = 0.1

now

mass = moles x molar mass

so

mass of KOH = 0.1 x 56.11

mass of KOH = 5.611

so

5.611 grams of solute is present

4)

a) mass of H20 = 1.5 x 18 = 27

b) mass of 02 = 0.5 x 32 = 16

c) mass of C6H1206 = 0.2 x 180 = 36

d) we know that

moles = number of gold atoms / 6.023 x 10^23

so

moles of Au = 10^9 / 6.023 x 10^23 = 1.66 x 10-15

now

mass = 1.66 x 10-15 x 197

mass of Au = 3.27 x 10-13 g

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