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3 years ago

The reaction of hydrogen gas with oxygen gas is what type of reaction precipitation single replacement

1 answer:

5 0

The reaction of hydrogen gas (H2) with oxygen gas (O2) is a COMBINATION or SYNTHESIS reaction, because multiple substances combine to form fewer substances. Here, the two gases form one substance, water (H2O):
2H2 + O2 -- > 2H2O

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When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular

Luda [366]

Answer: The empirical and molecular formula for the given organic compound is $C_2H_2O_4$ and $C_6H_4O_2$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=1.03g$

Mass of $H_2O=0.14g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.03 g of carbon dioxide, $\frac{12}{44}\times 1.03=0.28g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.14 g of water, $\frac{2}{18}\times 0.14=0.016g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00775 moles.

For Carbon = $\frac{0.023}{0.00775}=2.96\approx 3$

For Hydrogen = $\frac{0.016}{0.00775}=2.06\approx 2$

For Oxygen = $\frac{0.00775}{0.00775}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 2 : 1

Hence, the empirical formula for the given compound is $C_3H_{2}O_1=C_3H_2O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

$n=\frac{108.10g/mol}{54g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2$

Thus, the empirical and molecular formula for the given organic compound is $C_3H_2O$ and $C_6H_4O_2$

6 0

3 years ago

What charge does a proton have

Nata [24]

It has a positive charge of 1

5 0

3 years ago

Write chemical equations for the following electrolytes dissolving in water

Elden [556K]

$1)NaCl=>Na^{+}+Cl^{-}\\2)PbSO_{4}=>Pb^{2+}+SO_(4)^{2-}$

6 0

2 years ago

The specific heat capacity of concrete is 0.880 J/g °C

nata0808 [166]

Answer:

Solution given:

heat[Q]=?

temperature [T]=0.64°C

specific heat capacity [c]=0.880 J/g °C

mass[m]=3g

we have

Q=mcT=3*0.880*0.64°=1.69Joule

the required heat is 1.69Joule.

5 0

3 years ago

Read 2 more answers

How many grams of \text{NaCl}NaClstart text, N, a, C, l, end text will be produced from 18.0 \text{ g}18.0 g18, point, 0, start

Nezavi [6.7K]

Answer:

18.7887 g of NaCl

Explanation:

The question reads - How many grams of NaCl will be produced from 18.0 g of Na and 23.0 g of Cl2?

Let us start by writing out the balanced equation of the reaction:

Na + Cl2 ---> NaCl2

1 mole, each of Na and Cl2 is required to produce 1 mole of NaCl.

mole = mass/molar mass

Therefore

18 g of Na = 18/23 = 0.7826 mole

23 g of Cl2 = 23/71 = 0.3239 mole

In this case, the Na is in excess and the Cl2 becomes the limiting reagent. Hence

0.3239 mole of Cl2 will react with 0.3239 mole of Na to yield 0.3239 mole of NaCl.

mass of 0.3239 mole NaCl = 0.3239 x 58 = 18.7887 g

Hence, 18.7887 grams of NaCl will be produced from 18.0 g of Na and 23.0 g of Cl2.

3 0

3 years ago