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3 years ago

According to the Law of Conservation of Mass, how much water is produced if 22.5 g of

Chemistry

1 answer:

nadezda [96]3 years ago

4 0

Answer: The amount of water produced is 9.3 grams

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

$CH_4+2O_2\rightarrow CO_2+2H_2O$

mass of reactants = mass of methane + mass of oxygen = 22.5 g + 35.7 g = 58.2 g

mass of products = mass of carbon dioxide + mass of water = 48.9 g + mass of water

48.9 g + mass of water = 58.2 g

mass of water = 9.3 g

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3 years ago

Calculate the number of grams of CO that can react with 0.400 kg of Fe2O3.

aleksandr82 [10.1K]

Answer:

mass of CO = 210.42 g

mass in three significant figures = 210. g

Explanation:

Given data:

mass of Fe2O3 = 0.400 Kg

mass of CO= ?

Solution:

chemical equation:

Fe2O3 + 3CO → 2Fe + 3CO2

Now we will calculate the molar mass of Fe2O3 and CO.

Molar mass of Fe2O3 = (55.845 × 2) + (16 × 3) = 159.69 g/mol

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now we will convert the kg of Fe2O3 in g.

mass of Fe2O3 = 0.400 kg × 1000 = 400 g

number of moles of Fe2O3 = 400 g/ 159.69 g/mol = 2.505 mol

mass of CO = moles of Fe2O3 × 3( molar mass of CO)

mass of CO = 2.505 mol × 84 g/mol

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mass in three significant figures = 210. g

5 0

3 years ago

Read 2 more answers

I NEED HELP PLEASE, THANKS! :)

marin [14]

Answer:

$\large \boxed{1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2 }}$

Explanation:

1. Calculate the moles of copper(II) hydroxide

$\text{Moles of Cu(OH)}_{2} = \text{23.45 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \\\\\text{0.240 36 mol Cu(OH)}_{2}$

2. Calculate the molecules of copper(II) hydroxide

$\text{No. of molecules} = \text{0.240 36 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= 1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2}\\\text{The sample contains $\large \boxed{\mathbf{1.447 \times 10^{23}}\textbf{ molecules Cu(OH)}_{\mathbf{2}}}$}$