A natural resource from the Earth that exists in limited Supply

A certain alcohol contains only three elements, carbon, hydrogen, and oxygen. Combustion of a 60.00 gram sample of the alcohol p

marin [14]

Answer:

$C_2H_6O$

Explanation:

The first step is the calculation of the moles of $H_2O$ and $CO_2$ , so:

$114.6~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=2.6~mol~of~CO_2$

$70.44~g~H_2O\frac{1~mol~H_2O}{18~g~H_2O}=~3.9~mol~H_2O$

Now, in 1 mol of CO2 we have 1 mol of C and in 1 mol of $H_2O$ we have 1 mol of H. Additionally, if we want to calculate the moles of oxygen we need to calculate the grams of C and O and then do the substraction form the initial amount, so:

$2.6~mol~CO_2\frac{1~mol~C}{1~mol~CO_2}\frac{12~g~C}{1~mol~C}=31.25~g~of~C$

$3.9~mol~H_2O\frac{2~mol~H}{1~mol~H_2O}\frac{1~g~H}{1~mol~H}=7.82~g~of~H$

$Total~grams=~31.25~+~7.82=39.08~g$

$grams~of~O=60.00~g-~39.08~g=20.92~g~of~O$

Now we can convert the grams of O to moles, so:

$20.92~g~of~O\frac{1~mol~O}{16~g~O}=1.30~mol~O$

The next step is to divide all the mol values by the smallest one:

$O=\frac{1.30~mol~O}{1.30~mol~O}=~1$

$C=\frac{2.6~mol~C}{1.30~mol~O}=~2$

$H=\frac{7.82~mol~H}{1.30~mol~O}=6$

Therefore the formula is $C_2H_6O$

6 0

2 years ago

Lithium nitride reacts with water to produce ammonia and lithium hydroxide according to the equation, Li3N(s) + 3H2O (L) -->

Gemiola [76]

Answer:

0.480 grams

Explanation:

Li₃N(s) + 3D₂O (L) --------------------------> ND₃(g) + 3LiOD (aq)

1 : 3 : 1 : 3

Number of moles (n) = Mass in gram/ Molar Mass

Mass of ND₃ = 160 mg

= 0.16 g

Molar mass of ND₃= [14 + (3 x 2.014 )]

= 14 + 6.042

= 20.042 g/mol

Number of moles of ND₃ = 0.16/20.042

= 0.007983 moles

From the reaction equation, the mole ratio between Heavy water (D₂O ) and ND₃ is 3: 1.

This implies that the number of moles of Heavy water (D₂O ) required

= 3 x 0.007983 moles

= 0.023949 moles

Molar mass of Heavy water (D₂O )= [(2.014 x 2) + 16]

= 20.028 g/mol

Mass in grams of Heavy water (D₂O )= Number of moles x Molar mass

= 0.023949 x 20.028

= 0.4797 grams

≈ 0.480 grams

3 0

2 years ago

5.Are pure solids included in equilibrium expressions? Explain your answer.

tatiyna

Answer:

Pure solids or liquids are excluded from the equilibrium expression because their effective concentrations stay constant throughout the reaction. The concentration of a pure liquid or solid equals its density divided by its molar mass.

Explanation:

Hope this helps!!!

8 0

2 years ago

An

Basile [38]

Answer:

The answer of this question is molecule

4 0

2 years ago

How many number of moles are present in 200. g of sodium bicarbonate (NaHCO3)

nikdorinn [45]

You can solve this problem through dimensional analysis.

First, find the molar mass of NaHCO3.

Na = 22.99 g

H = 1.008 g

C = 12.01 g

O (3) = 16 (3) g

Now, add them all together, you end with with the molar mass of NaHCO3.

22.99 + 1.008 + 12.01 + 16(3) = 84.008 g NaHCO3. This number means that for every mole of NaHCO3, there is 84.008 g NaHCO3. In simpler terms, 1 mole NaHCO3 = 84.008 g NaHCO3.

After finding the molar mass of sodium bicarbonate, now you can use dimensional analysis to solve for the number of moles present in 200. g of sodium bicarbonate.

$200. g NaHCO_3 * \frac{1 mole NaHCO_3}{84.008 g NaHCO_3}$

Cross out the repeating units which are g NaHCO3, and the remaining unit is mole NaHCO3

200. * 1 = 200

200/ 84.008 = 2.38

Notice how there are only 3 sig figs in the answer. This is because the given problem only gave three sig figs.

Your final answer is 2.38 mol NaHCO3.

8 0

2 years ago