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2 years ago

What mass of Cu(s) would be produced if 0.40 mol of Cu2O(s) was reduced completely with excess H2(g) ?

Chemistry

1 answer:

yan [13]2 years ago

4 0

Answer:

50.8g

Explanation:

Given parameters:

Nubmer of mole of Cu₂O = 0.4mol

Unknown:

Mass of Cu produced = ?

Solution:

The balanced reaction equation:

Cu₂O + H₂ ⇄ 2Cu + H₂O

From the reaction, we know that:

1 mole of Cu₂O produced 2 moles of Cu

0.4 mol of Cu₂O would give 0.8mol of Cu

Now we know the number of moles of Cu produced because the limiting reactant here is Cu₂O. It was used up in the reaction in which the hydrogen gas was in excess.

To find the mass of Cu produced, we use the equation below :

Mass of Cu = Number of moles of Cu x molar mass of Cu

Molar mass of Cu = 63.5gmol⁻¹

Mass of Cu = 0.8 x 63.5 = 50.8g

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if 5.50 mol of calcium carbide (CaC 2 ) reacts with an excess of water, how many moles of acetylene (C 2 H 2 ) , a gas used in w

Leona [35]

Answer:

5.5moles

Explanation:

CaC2 + 2H2O —> Ca(OH)2 + C2H2

From the equation, the following were observed:

1mole of CaC2 reacted to produced 1mol of C2H2.

Therefore, 5.5moles of CaC2 will also produce 5.5moles of C2H2

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2 years ago

5. Rubbing alcohol is a commonly used disinfectant and has a cooling effect when applied to the skin. The active ingredient in r

WITCHER [35]

Answer:

Explanation:

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2 years ago

. A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5% and 15%. A mixture

postnew [5]

Answer:

A) Mass flow rate of air = 22.892 kmol/hr

B)percentage by mass of oxygen in the product gas = 22.52%

Explanation:

We are given that the mixture containing 9.0 mole% methane in air flowing.

Thus, we have 0.09 mole of methane(CH4) and the remaining will be the air which is (100% - 9%) = 91% = 0.91

Molar mass of CH4 = 12 + 1(4) = 16 g/mol

We are given the average molecular weight of air = 29 g/mol

Thus;

Average molar mass of air and methane mixture is;

M_avg = (0.09 × 16) + (0.91 × 29)

M_avg = 27.83 g/mol

We are told that air flowing at a rate of 7 × 10² kg/h = 700 kg/h

Thus;

Mass flow rate of CH4 in air mixture = 700kg/h × (0.09CH4)/1 mix × (1/27.83kg/kmol) = 2.264 kmol/hr

Mass flow rate of air in mixture = 2.264kmol/h × 0.91kmol air/0.09kmolCH4 = 22.892 kmol/hr

We are told that the mixture is capable of being ignited if the mole percent of methane is between 5% and 15%.

Thus, for 5% of methane, the air required will be;

2.264kmol/h × 0.95kmol air/0.05kmol CH4 = 43.016 kmol/hr

Now, the dilution air needed will be =

43.016 - 22.892 = 20.124 kmol/hr

Total mass flow rate of mixture =

700kg/hr + (20.124kmol/hr × 29kg/mol) = 1283.596 kg/hr

We are told that air consist of 21 mole% Oxygen (O2).

Molar mass of oxygen = 32

Thus;

Mass fraction of oxygen in the product gas = 43.016kmol/h × (0.21molO2/1mol air) × (32kg oxygen/1kmol oxygen) × (1/1283.596kg/h) = 0.2252

Thus, written in percentage form, we have; 22.52%

So, percentage by mass of oxygen in the product gas = 22.52%

3 0

2 years ago

The molar mass of nitrogen (N2) is 28.02 g/mol. What is the mass, in grams, of 4.60 mol of N2?

kotegsom [21]

28.02g ------ 1 mol
X g ------------ 4.60 mol

X = (28.02×4.60)/1
X = 128,892 g

7 0

2 years ago

Read 2 more answers

The balanced equation shows how sodium chloride reacts with silver nitrate to form sodium nitrate and silver chloride. NaCl + Ag

Verdich [7]

<span>NaCl First calculate the molar mass of NaCl and AgNO3 by looking up the atomic weights of each element used in either compound Sodium = 22.989769 Chlorine = 35.453 Silver = 107.8682 Nitrogen = 14.0067 Oxygen = 15.999 Now multiply the atomic weight of each element by the number of times that element is in each compound and sum the results For NaCl 22.989769 + 35.453 = 58.44277 For AgNO3 107.8682 + 14.0067 + 3 * 15.999 = 169.8719 Now calculate how many moles of each substance by dividing the total mass by the molar mass For NaCl 4.00 g / 58.44277 g/mol = 0.068443 mol For AgNO3 10.00 g / 169.8719 g/mol = 0.058868 Looking at the balanced equation for the reaction, there is a 1 to 1 ratio in molecules for the reaction. Since there is a smaller number of moles of AgNO3 than there is of NaCl, that means that there will be some NaCl unreacted, so the excess reactant is NaCl</span>

8 0

2 years ago