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Yuri [45]
4 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]4 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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A student needs to prepare 50.0 mL of a 1.20 M aqueous H2O2 solution. Calculate the volume of 4.9 M H2O2 stock solution that sho
Nonamiya [84]

Answer : The volume of 4.9 M H_2O_2 stock solution used to prepare the solution is, 12.24 ml

Solution : Given,

Molarity of aqueous H_2O_2 solution = 1.20 M = 1.20 mole/L

Volume of aqueous H_2O_2 solution = 50.0 ml = 0.05 L

(1 L = 1000 ml)

Molarity of H_2O_2 stock solution = 4.9 M = 4.9 mole/L

Formula used :

M_1V_1=M_2V_2

where,

M_1 = Molarity of aqueous H_2O_2 solution

M_2 = Molarity of H_2O_2 stock solution

V_1 = Volume of aqueous H_2O_2 solution

V_2 = Volume of H_2O_2 stock solution

Now put all the given values in this formula, we get the volume of H_2O_2 stock solution.

(1.20mole/L)\times (0.05L)=(4.9mole/L)\times V_2

By rearranging the term, we get

V_2=0.01224L=12.24ml

Therefore, the volume of 4.9 M H_2O_2 stock solution used to prepare the solution is, 12.24 ml

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A solution is made by dissolving 3.35 g of fructose in 35.0 ml of water. What is the molality of fructose in the solution? Assum
ludmilkaskok [199]

Answer:

m = 0.531 molal

Explanation:

∴ m fructose = 3.35 g

∴ V water = 35.0 mL

∴ ρ H2O = 1 g/mL

  • molality = moles solute / Kg solvent

∴ Mw fructose = 180.16 g/mol

⇒ moles fructose = 3.35 g * ( mol / 180.16 g) = 0.0186 mol fructose

⇒ m H2O = 35.0 mL * ( 1 g/mL ) * ( Kg/1000g) = 0.035 Kg H2O

⇒ molality (m) = 0.0186 mol fructose / 0.035 Kg H2O

⇒ m = 0.531 molal

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3 years ago
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