All categories

2 years ago

How many moles of sodium nitrate NaNO3 would be produced from the complete reaction of 253 g sodium chromate Na2CrO4

Chemistry

2 answers:

Elenna [48]2 years ago

7 0

Answer: 3.12

Hope you pass

Stolb23 [73]2 years ago

5 0

<h3><u>Answer;</u></h3>

= 3.12 moles of NaNO3

<h3><u>Explanation</u>;</h3>

The equation for the reaction is;

Pb(NO3)2+Na2CrO4 → PbCrO4+2NaNO3

Molar ratios in the reaction are ; 1 : 1 : 1 : 2

Which means; One mole of (Na2CrO4) produces two moles of (NaNO3)

We can calculate the number of moles of Na2CrO4,

Moles (na2CrO4) = 253g /162

= 1.56 moles

Moles of NaNO3 will be;

= 1.56 moles x 2 = 3.12 moles

Therefore, the moles of NaNO3 that will be produced will be 3.12 moles

You might be interested in

8. Explain how Elianto oil is obtained from maize seeds.

Mashutka [201]

Answer:

Almost all corn oil is expeller-pressed

Explanation:

then solvent-extracted using hexane or 2-methylpentane (isohexane). The solvent is evaporated from the corn oil, recovered, and re-used. After extraction, the corn oil is then refined by degumming and/or alkali treatment, both of which remove phosphatides.Oct 16, 2020

3 0

1 year ago

Find the number of moles of water that can be formed if you have 170 mol of hydrogen gas and 80 mol of oxygen gas. Express your

Sav [38]

<u>Answer:</u> The amount of water that can be formed is 160 moles

<u>Explanation:</u>

We are given:

Moles of hydrogen gas = 170 moles

Moles of oxygen gas = 80 moles

The chemical equation for the reaction of hydrogen gas and oxygen gas follows:

$2H_2+O_2\rightarrow 2H_2O$

By Stoichiometry of the reaction:

1 mole of oxygen gas reacts with 2 moles of hydrogen gas

So, 80 moles of oxygen gas will react with = $\frac{2}{1}\times 80=160mol$ of hydrogen gas

As, given amount of hydrogen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of oxygen gas produces 2 moles of water

So, 80 moles of oxygen gas will produce = $\frac{2}{1}\times 80=160moles$ of water

Hence, the amount of water that can be formed is 160 moles

3 0

2 years ago

When 13.6 g of calcium chloride, CaCl2, was dissolved in 100.0 mL of water in a coffee cup calorimeter, the temperature rose fro

DanielleElmas [232]

Answer:

THE ENTHALPY OF SOLUTION IS 3153.43 J/MOL OR 3.15 KJ/MOL.

Explanation:

1. write out the variables given:

Mass of Calcium chloride = 13.6 g

Change in temperature = 31.75°C - 25.00°C = 6.75 °C

Density of the solution = 1.000 g/mL

Volume = 100.0 mL = 100.0 mL

Specific heat of water = 4.184 J/g °C

Mass of the water = unknown

2. calculate the mass of waterinvolved:

We must first calculate the mass of water in the bomb calorimeter

Mass = density * volume

Mass = 1.000 * 100

Mass = 0.01 g

3. calculate the quantity of heat evolved:

Next is to calculate the quantity of heat evolved from the reaction

Heat = mass * specific heat of water * change in temperature

Heat = mass of water * specific heat *change in temperature

Heat = 13.6 g * 4.184 * 6.75

Heat = 13.6 g * 4.184 J/g °C * 6.75 °C

Heat = 384.09 J

Hence, 384.09J is the quantity of heat involved in the reaction of 13.6 g of calcium chloride in the calorimeter.

4. calculate the molar mass of CaCl2:

Next is to calculate the molar mas of CaCl2

Molar mass = ( 40 + 35.5 *2) = 111 g/mol

The number of moles of 13.6 g of CaCl2 is then:

Number of moles of CaCl2 = mass / molar mass

Number of moles = 13.6 g / 111 g/mol

Number of moles = 0.1225 mol

So 384.09 J of heat was involved in the reaction of 1.6 g of CaCl2 in a calorimter which translates to 0.1225 mol of CaCl2..

5. Calculate the enthalpy of solution in kJ/mol:

If 1 mole of CaCl2 is involved, the heat evolved is therefore:

Heat per mole = 384.09 J / 0.1225 mol

Heat = 3 135.43 J/mol

The enthalpy of solution is therefore 3153.43 J/mol or 3.15 kJ/mol.

5 0

2 years ago

A student assistant is cleaning up after a chemistry laboratory exercise and finds three one-liter bottles containing alcohol so

Svetllana [295]

The approximate alcohol content is 210 ml.

Explanation:

It can be deduced from the question that each bottle is of 1000ml or 1 litre.

The first bottle is one half full means it has 500 ml of solution and it has 20% alcohol in it. So volume of alcohol in the solution is

20/100*500

=100 ml

The first bottle is one fifth full, so the volume of mixture is 1/5th of 1000ml

so it is 200ml having 30% alcohol

30/100*200

= 60 ml

The third bottle is one tenth full so its volume is 1/10*1000

100 ml. having 50% of alcohol

50/100*100

50 ml.

The alcohol content obtained from all these 3 litres is:

100+60+50

= 210 ml of alchohol is obtained from 800 ml of mixture.

3 0

2 years ago

1) Aluminum sulphate can be made by the following reaction: 2AlCl3(aq) + 3H2SO4(aq) Al2(SO4)3(aq) + 6 HCl(aq) It is quite solubl

kolezko [41]

Answer:

88.9%

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2AlCl3(aq) + 3H2SO4(aq) —> Al2(SO4)3(aq) + 6HCl(aq)

Step 2:

Determination of the masses of AlCl3 and H2SO4 that reacted and the mass of Al2(SO4)3 produced from the balanced equation.

Molar mass of AlCl3 = 27 + (35.5x3) = 133.5g/mol

Mass of AlCl3 from the balanced equation = 2 x 133.5 = 267g

Molar mass of H2SO4 = (2x1) + 32 + (16x4) = 98g/mol

Mass of H2SO4 from the balanced equation = 3 x 98 = 294g

Molar mass of Al2(SO4)3 = (27x2) + 3[32 + (16x4)]

= 54 + 3[32 + 64]

= 54 + 3[96] = 342g/mol

Mass of Al2(SO4)3 from the balanced equation = 1 x 342 = 342g

Summary:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4 to produce 342g of Al2(SO4)3.

Step 3:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4.

Therefore, 25g of AlCl3 will react with = (25 x 294)/267 = 27.53g of H2SO4.

From the calculations made above, we see that only 27.53g out 30g of H2SO4 given were needed to react completely with 25g of AlCl3.

Therefore, AlCl3 is the limiting reactant and H2SO4 is the excess.

Step 4:

Determination of the theoretical yield of Al2(SO4)3.

In this case we shall be using the limiting reactant because it will produce the maximum yield of Al2(SO4)3 since all of it is used up in the reaction.

The limiting reactant is AlCl3 and the theoretical yield of Al2(SO4)3 can be obtained as follow:

From the balanced equation above,

267g of AlCl3 reacted to produce 342g of Al2(SO4)3.

Therefore, 25g of AlCl3 will react to produce = (25 x 342) /267 = 32.02g of Al2(SO4)3.

Therefore, the theoretical yield of Al2(SO4)3 is 32.02g

Step 5:

Determination of the percentage yield of Al2(SO4)3.

This can be obtained as follow:

Actual yield of Al2(SO4)3 = 28.46g

Theoretical yield of Al2(SO4)3 = 32.02g

Percentage yield of Al2(SO4)3 =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 28.46/32.02 x 100

Percentage yield = 88.9%

Therefore, the percentage yield of Al2(SO4)3 is 88.9%

3 0

2 years ago