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3 years ago

How many moles of H 2 can be formed if a 3.24 g sample of M g reacts with excess H C l ?

Chemistry

1 answer:

Ganezh [65]3 years ago

3 0

Answer:

0.135 mole of H2.

Explanation:

We'll begin by calculating the number of mole in 3.24 g of Mg. This can be obtained as follow:

Mass of Mg = 3.24 g

Molar mass of Mg = 24 g/mol

Mole of Mg =?

Mole = mass /Molar mass

Mole of Mg = 3.24/24

Mole of Mg = 0.135 mole

Next, we shall write the balanced equation for the reaction. This is illustrated below:

Mg + 2HCl —> MgCl2 + H2

From the balanced equation above,

1 mole of Mg reacted to produce 1 mole of H2.

Finally, we shall determine the number of mole of H2 produced by reacting 3.24 g (i.e 0.135 mole) of Mg. This can be obtained as follow:

From the balanced equation above,

1 mole of Mg reacted to produce 1 mole of H2.

Therefore, 0.135 mole of Mg will also react to produce 0.135 mole of H2.

Thus, 0.135 mole of H2 can be obtained from the reaction.

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3 years ago

Jill is making lemonade. She mixes together some water, lemon juice, and sugar. The lemonade is classified as a(n) _________. A)

PolarNik [594]

Hello, Thanks for making your post on brainly!

I would say the best answer is C)Mixture

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Hope this helps. Have a great day.

8 0

2 years ago

Read 2 more answers

Use this equation for the following problems: 2NaN3 --> 2Na+3N2

olchik [2.2K]

Answer:

1) 65.0

2) 16.434 L = 16434 mL.

Explanation:

2NaN₃ → 2Na + 3N₂,

It is clear from the balanced equation that 2.0 moles of NaN₃ are decomposed to 2.0 moles of Na and 3.0 moles of N₂.

Q1: How many grams of NaN₃ are needed to make 23.6L of N₂?

Density of N₂ = 0.92 g/L which means that every 1.0 L of N₂ contains 0.92 g of N₂.

Now, we can get the mass of N₂ in 23.6 L N₂ using cross multiplication:

1.0 L of N₂ contains → 0.92 g of N₂.

23.6 L of N₂ contains → ??? g of N₂.

∴ The mass of N₂ in 23.6 L of N₂ = (23.6 L)(0.92 g)/(1.0 L) = 21.712 g.

We can get the no. of moles of 23.6 L of N₂ (21.712 g) using the relation:

n = mass/molar mass = (21.712 g)/(28.0 g/mol) = 0.775 mol.

We can get the no. of moles of NaN₃ needed to produce 0.775 mol of N₂:

using cross multiplication:

2.0 moles of NaN₃ produce → 3.0 moles of N₂, from the balanced equation.

??? mol of NaN₃ produce → 0.775 moles of N₂.

∴ The no. of moles of NaN₃ needed = (2.0 mol)(0.775 mol)/(3.0 mol) = 0.517 mol.

Finally, we can get the grams of NaN₃ needed:

mass = no. of moles x molar mass = (0.517 mol)(65.0 g/mol) = 33.6 g.

Q2: How many mL of N₂ result if 8.3 g Na are also produced?

We need to get the no. of moles of 8.3 g Na using the relation:

n = mass/atomic mass = (8.3 g)/(22.98 g/mol) = 0.36 mol.

We can get the no. of moles of N₂ produced with 0.36 mol of Na:

using cross multiplication:

2.0 moles of Na produced with → 3.0 moles of N₂, from the balanced equation.

0.36 moles of Na produced with → ??? moles of N₂.

∴ The no. of moles of N₂ needed = (3.0 mol)(0.36 mol)/(2.0 mol) = 0.54 mol.

We can get the mass of 0.54 mol of N₂:

mass = no. of moles x molar mass = (0.54 mol)(28.0 g/mol) = 15.12 g.

Now, we can get the mL of 15.12 g of N₂:

using cross multiplication:

1.0 L of N₂ contains → 0.92 g of N₂, from density of N₂ = 0.92 g/L.

??? L of N₂ contains → 15.12 g of N₂.

∴ The volume of N₂ result = (1.0 L)(15.12 g)/(0.92 g) = 16.434 L = 16434 mL.

4 0

2 years ago

The molarity of an aqueous sodium phosphate solution is 0.650 M. What is the molality of sodium ions present in this solution? T

netineya [11]

Answer:

molality of sodium ions is 1.473 m

Explanation:

Molarity is moles of solute per litre of solution

Molality is moles of solute per kg of solvent.

The volume of solution = 1 L

The mass of solution = volume X density = 1000mL X 1.43 = 1430 grams

The mass of solute = moles X molar mass of sodium phosphate = 0.65X164

mass of solute = 106.6 grams

the mass of solvent = 1430 - 106.6 = 1323.4 grams = 1.3234 Kg

the molality = $\frac{moles of solute}{mass of solvent in kg}=\frac{0.65}{1.323}= 0.491m$

Thus molality of sodium phosphate is 0.491 m

Each sodium phosphate of molecule will give three sodium ions.

Thus molality of sodium ions = 3 X 0.491 = 1.473 m

7 0

3 years ago

In the industrial synthesis of hydrogen, mixtures of CO and H2 are enriched in H2 by allowing the CO to react with steam. The ch

saw5 [17]

Answer:

CO(g) + H2(g) + H2O(g) ==> CO2(g) + 2H2(g)

Explanation:

In the industry, hydrogen is prepared from water and hydrocarbons. Water gas being the major method of preparation of hydrogen industrially.

The water-gas reaction is an industrial process in which steam is passed over red-hot coke giving a gaseous mixture of carbon monoxide and hydrogen:

C + H2O(g) → CO + H2.

The mixture of CO and H2 is Futher passed through steam according to the equation:

CO(g) + H2(g) + H2O(g) ==> CO2(g) + 2H2(g) to give hydrogen and carbon dioxide.

8 0

3 years ago