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3 years ago

If the density of water is 1g/ml, what is the mass of 254ml of water?

1 answer:

8 0

The mass of 254 mL of water is 254 g. Since the density of water is 1g/mL, we can simply multiply the density 1g/mL by 254 mL of water and get 254 g as our answer. Since mL is in the numerator and denominator, mL cancels out and we are left with g only.

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Consider the reaction Mg(s) + I2 (s) → MgI2 (s) Identify the limiting reagent in each of the reaction mixtures below:

Lapatulllka [165]

Answer:

a) Nor Mg, neither I2 is the limiting reactant.

b) I2 is the limiting reactant

c) Mg is the limiting reactant

d) Mg is the limiting reactant

e) Nor Mg, neither I2 is the limiting reactant.

f) I2 is the limiting reactant

g) Nor Mg, neither I2 is the limiting reactant.

h) I2 is the limiting reactant

i) Mg is the limiting reactant

Explanation:

Step 1: The balanced equation:

Mg(s) + I2(s) → MgI2(s)

For 1 mol of Mg we need 1 mol of I2 to produce 1 mol of MgI2

a. 100 atoms of Mg and 100 molecules of I2

We'll have the following equation:

100 Mg(s) + 100 I2(s) → 100MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

b. 150 atoms of Mg and 100 molecules of I2

We'll have the following equation:

150 Mg(s) + 100 I2(s) → 100 MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 100 Mg atoms. There will remain 50 Mg atoms.

There will be produced 100 MgI2 molecules.

c. 200 atoms of Mg and 300 molecules of I2

We'll have the following equation:

200 Mg(s) + 300 I2(s) →200 MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 200 I2 molecules. There will remain 100 I2 molecules.

There will be produced 200 MgI2 molecules.

d. 0.16 mol Mg and 0.25 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.16 mol of I2. There will remain 0.09 mol of I2.

There will be produced 0.16 mol of MgI2.

e. 0.14 mol Mg and 0.14 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.14 mol of Mg and 0.14 mol of I2. there will be produced 0.14 mol of MgI2

f. 0.12 mol Mg and 0.08 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.08 moles of Mg. There will remain 0.04 moles of Mg.

There will be produced 0.08 moles of MgI2.

g. 6.078 g Mg and 63.455 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 6.078 grams / 24.31 g/mol = 0.250 moles

Number of moles I2 = 63.455 grams/ 253.8 g/mol = 0.250 moles

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.250 mol of Mg and 0.250 mol of I2. there will be produced 0.250 mol of MgI2

h. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 2.00 grams/ 253.8 g/mol = 0.00788 moles

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.00788 moles of Mg. There will remain 0.03322 moles of Mg.

There will be produced 0.00788 moles of MgI2.

i. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 20.00 grams/ 253.8 g/mol = 0.0788 moles

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.0411 moles of Mg. There will remain 0.0377 moles of I2.

There will be produced 0.0411 moles of MgI2.

4 0

3 years ago

Oxidation state of Nitrogen in N2O5

balandron [24]

Explanation:

Oxidation state of Nitrogen in N2O5 is +5

7 0

1 year ago

Calculate the molality of 75.0 grams of MgCl2 (molar mass=95.21 g/mol) dissolved in 500.0 g of solvent.

nordsb [41]

Answer: The molality of magnesium chloride is 1.58 m

Explanation:

To calculate the molality of solution, we use the equation:

$\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

Where,

$m_{solute}$ = Given mass of solute (magnesium chloride) = 75.0

$M_{solute}$ = Molar mass of solute (magnesium chloride) = 95.21 g/mol

$W_{solvent}$ = Mass of solvent = 500.0 g

Putting values in above equation, we get:

$\text{Molality of }MgCl_2=\frac{75.0\times 1000}{95.21\times 500.0}\\\\\text{Molality of }MgCl_2=1.58m$

Hence, the molality of magnesium chloride is 1.58 m

5 0

3 years ago

Read 2 more answers

You dissolve 8.65 grams of lead(l) nitrate in water and then you add 2 50 grams of aluminum. This reaction occurs 2AI(S)+ 3Pb(NO

olga55 [171]

Answer: The theoretical yield of solid lead comes out to be 5.408 grams.

Explanation:

To calculate the moles, we use the following equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of Lead nitrate:

Given mass of lead nitrate = 8.65 grams

Molar mass of lead nitrate = 331.2 g/mol

Putting values in above equation, we get:

$\text{Number of moles}=\frac{8.65g}{331.2g/mol}=0.0261moles$

Moles of Aluminium:

Given mass of aluminium = 2.5 grams

Molar mass of aluminium = 27 g/mol

Putting values in above equation, we get:

$\text{Number of moles}=\frac{2.5g}{27g/mol}=0.0925moles$

For the given chemical reaction, the equation follows:

$2AI(s)+3Pb(NO_3)_2(aq.)\rightarrow 3Pb(s)+2AI(NO3)_3(aq.$

By Stoichiometry:

3 moles of lead nitrate reacts with 2 moles of aluminium

So, 0.0261 moles of lead nitrate are produced by = $\frac{2}{3}\times 0.0261=0.0174moles$ of aluminium.

As, the required amount of aluminium is less than the given amount. Hence, it is considered as the excess reagent.

Lead nitrate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of lead nitrate are produces 3 moles of lead metal.

So, 0.0261 moles of lead nitrate will produce = $\frac{3}{3}\times 0.0261=0.0261moles$ of lead metal.

Now, to calculate the grams or theoretical yield of lead metal, we put in the mole's equation, we get:

Molar mass of lead = 207.2 g/mol

Putting values in above equation, we get:

$0.0261mol=\frac{\text{Given mass}}{207.2g/mol}$

Mass of lead = 5.408 grams

Hence, the theoretical yield of solid lead comes out to be 5.408 grams.

8 0

3 years ago

What does changing densities cause to occur

Dafna1 [17]

Answer:

Explanation:

Please Elaborate

8 0

3 years ago