Copper(i) oxide, cu2o, is reduced to metallic copper by heating in a stream of hydrogen gas. what mass of water is produced when 10.00 g copper is formed?

Answers

Answer 1

Answer is: mass of water is 1.41 g.
Balanced chemical reaction: Cu₂O + H₂ → 2Cu + H₂O.
m(Cu) = 10.00 g.
n(Cu) = m(Cu) ÷ M(Cu).
n(Cu) = 10 g ÷ 63.55 g/mol.
n(Cu) = 0.157 mol.
From chemical reaction: n(Cu) : n(H₂O) = 2 : 1.
n(H₂O) = 0.079 mol.
m(H₂O) = n(H₂O) · M(H₂O).
m(H₂O) = 0.079 mol · 18 g/mol.
m(H₂O) = 1.41 g.

Answer 2

Final answer:

The mass of water produced when 10.00 g of copper is formed by the reduction of Cu2O with hydrogen gas is 1.4131 g. This is calculated using stoichiometry and the molar mass of water.

Explanation:

The mass of water produced when 10.00 g of copper is formed from the reduction of copper(I) oxide (Cu2O) by hydrogen gas (H2) can be determined by using stoichiometry. The reaction is as follows:

Cu2O(s) + H2(g) → 2 Cu(s) + H2O(g)

First, calculate the moles of copper produced using its molar mass. Since the copper produced is 10.00 g, and the molar mass of copper is approximately 63.55 g/mol, the moles of copper formed are:

10.00 g Cu × (1 mol Cu / 63.55 g Cu) = 0.157 mol Cu

According to the reaction, 1 mole of Cu2O produces 2 moles of Cu, so we have:

0.157 mol Cu × (1 mol Cu2O / 2 mol Cu) = 0.0785 mol Cu2O

For each mole of Cu2O reduced, 1 mole of water (H2O) is produced:

0.0785 mol Cu2O × (1 mol H2O / 1 mol Cu2O) = 0.0785 mol H2O

Finally, to find the mass of water produced:

0.0785 mol H2O × (18.015 g H2O / 1 mol H2O) = 1.4131 g H2O

Therefore, the mass of water produced is 1.4131 g.

Related Questions

Is it possible for two yellow belied noombats to have a green bellied child?

Answers

If both of them has yellow as the dominant trait, then green may be recessive. In that case there is a chance for them to have a green bellied child.

Hope this helps :)

which of the following represents a stable octet? A 1s2 2s2 2p6 3s2 3p6 4s2
B [He] 2s2 2p3
C 1s2 2s2 2p1
D [Ne] 3s2 3p6

Answers

[Ne]3s^23p^6 ( answer D) represent a sample of an octet. Octet is the tendency of an atom to have eight electrons in the valence shell. [Ne]3s^2 3p^6 is an octet since its valence electron that is3s^23p^6 has eight electrons.

How many grams of glucose are needed to prepare 400. g of a 2.00% (m/m) glucose solution g?

Answers

Final answer:

To create 400 grams of a 2% mass/mass glucose solution, you need 8 grams of glucose.

Explanation:

You're trying to find out how many grams of glucose are needed to prepare a 400 gram 2% mass/mass glucose solution. A 2% w/w glucose solution means that for every 100 grams of solution, 2 grams are glucose. Therefore, if you have 400 grams of solution, the amount of glucose required will be 2% of 400 grams.

To calculate this, you will multiply 400 grams by 0.02 (which is the decimal equivalent of 2%). So, 400 grams * 0.02 = 8 grams. Therefore, you need 8 grams of glucose to prepare 400 grams of a 2% w/w glucose solution.

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He radioisotope radon-222 has a half-life of 3.8 days. how much of a 65-g sample of radon-222 would be left after approximately 15 days?

Answers

Given: Half life of Rn = 3.8 days
Therefore in 15 days, system crosses 15/3.8 = 3.94 ≈ 4 half-life

Now, Initial amount of Rn = 65 g

∴ Amount of Rn left after 4 half life i.e. 15 days = [tex] \frac{65}{2^4} [/tex] = 4.06 g

Nonmetals gain electrons under certain conditions to attain a noble-gas electron configuration. how many electrons must be gained by the element c?

Answers

System is said to have achieved noble-gas configuration, when it's valance shell is completely filled.

Atomic number of carbon is 6. Thus, it has 6 electrons.

The electronic configuration of carbon is 1s2 2s2 2p2

Now, the inert gas closest to C is Ne, whose atomic number is 10.

Thus, there are excess of 4 electrons in Ne as compared to C.

Hence, carbon must gain 4 electrons to achieve noble-gas configuration.

Alternatively, C can also lose 4 electron to achieve noble gas configuration of He.

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