Question

Скільки молів води та яка маса карбон(ІV) оксиду CO, утвориться в результаті згоряння 12 г метану СН ?

Answer 1

Объяснение:

Рівняння реакції: 2C2H6+4O2=4CO2+6H2O

n(C2H6) = 12/30 =0,4 моль

04/2 = х/4

х = 0,8 моль

V = 0,8 • 22,4 = 17,92 л

Answer 2

Calculation of Water and Carbon(IV) Oxide from the Combustion of Methane

To calculate the amount of water and carbon(IV) oxide formed from the combustion of 12 grams of methane (CH4), we can use the balanced chemical equation for the combustion of methane:

CH4 + 2O2 → CO2 + 2H2O

According to this equation, 1 mole of methane reacts with 2 moles of oxygen to produce 1 mole of carbon(IV) oxide and 2 moles of water.

Step 1: Calculate the Moles of Methane

The molar mass of methane (CH4) is 16.04 g/mol. Therefore, to find the moles of methane in 12 grams, we use the formula:

moles = mass / molar mass

So, moles of methane = 12 g / 16.04 g/mol

Step 2: Determine the Moles of Water and Carbon(IV) Oxide

Using the mole ratio from the balanced chemical equation, we can determine the moles of water and carbon(IV) oxide produced.

From the balanced equation, 1 mole of methane produces 2 moles of water and 1 mole of carbon(IV) oxide.

Step 3: Calculate the Mass of Water and Carbon(IV) Oxide

Finally, we can calculate the mass of water and carbon(IV) oxide produced using the moles obtained in the previous step and their respective molar masses.

I will now calculate the moles of methane, moles of water, moles of carbon(IV) oxide, and the mass of water and carbon(IV) oxide produced.