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Yuri [45]
3 years ago
8

During a combustion reaction, 12.2 grams of methane reacts with 14 g of oxygen. The reaction produces carbon dioxide and water.

If 20 grams of water are produced, how many grams of carbon dioxide are produced?
Chemistry
1 answer:
iogann1982 [59]3 years ago
5 0

The reaction produces 6 g of carbon dioxide.

The word equation is

methane + oxygen → carbon dioxide + water

 12.2 g    +    4 g     →            x g          +  20 g

The <em>Law of Conservation of Mass</em> tells us the total mass of the reactants must equal the mass of the products. Then

12.2 g + 14 g =<em> x</em> g + 20 g

<em>x</em> = 12.2 + 14 – 20 = 6

The mass of carbon dioxide is 6 g.

<em>Note</em>: The answer can have <em>no decimal places</em> because you gave none for the masses of oxygen and water.

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What is that the theoretical yield of aluminum oxide I if 3.20 mol of aluminum metal is exposed to 2.70 mole of oxygen
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Answer:

163.2g

Explanation:

First let us generate a balanced equation for the reaction. This is shown below:

4Al + 3O2 —> 2Al2O3

From the question given, were were told that 3.2moles of aluminium was exposed to 2.7moles of oxygen. Judging by this, oxygen is excess.

From the equation,

4moles of Al produced 2moles of Al2O3.

Therefore, 3.2moles of Al will produce = (3.2x2)/4 = 1.6mol of Al2O3.

Now, let us covert 1.6mol of Al2O3 to obtain the theoretical yield. This is illustrated below:

Mole of Al2O3 = 1.6mole

Molar Mass of Al2O3 = (27x2) + (16x3) = 54 + 48 =102g/mol

Mass of Al2O3 =?

Number of mole = Mass /Molar Mass

Mass = number of mole x molar Mass

Mass of Al2O3 = 1.6 x 102 = 163.2g

Therefore the theoretical of Al2O3 is 163.2g

8 0
3 years ago
30.0 ml of an hf solution were titrated with 22.15 ml of a 0.122 m koh solution to reach the equivalence point. what is the mola
a_sh-v [17]
Answer is: molarity of hydrofluoric solution is 0.09 M.

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V(HF) = 30.0 mL.
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From chemical reaction: n(HF) : n(KOH) = 1 : 1.
n(HF) = n(KOH).
c(HF) · V(HF) = c(KOH) · V(KOH).
c(HF) = c(KOH) · V(KOH) ÷ V(HF).
c(HF) = 0.122 M · 22.15 mL ÷ 30 mL:
c(HF) = 0.09 M.
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