CHEMISTRY COLLEGE

Calculate the volume of 0.100 m hcl required to neutralize 1.00 g of ba(oh)2 (molar mass = 171.3 g/mol).

Answers

Answer 1

Answer:

The balanced chemical equation between HCl and $Ba(OH)_(2)$ is:

$2HCl (aq) + Ba(OH)_(2)(aq) -->BaCl_(2)(aq) + 2 H_(2)O(l)$

Moles of $Ba(OH)_(2)$ = $1.00 g Ba(OH)_(2) * (1 mol Ba(OH)_(2))/(171.3 g Ba(OH)_(2)) = 0.00584 mol Ba(OH)_(2)$

Moles of HCl required to neutralize $Ba(OH)_(2)$ :

$0.00584 mol Ba(OH)_(2) * ( 2 mol HCl)/(1 mol Ba(OH)_(2)) = 0.01168 mol HCl$

Calculating the volume of HCl from moles and molarity:

$0.01168 mol HCl * (1 L)/(0.100 mol) * (1000 mL)/(1 L) = 116.8 mL$

Answer 2

Answer:

Answer:- 117 mL of HCl are used.

Solution:- The balanced equation for the reaction of HCl with barium hydroxide is written as:

$2HCl(aq)+Ba(OH)_2(aq)\rightarrow BaCl_2(aq)+2H_2O(l)$

From above equation, HCl and $Ba(OH)_2$ react in 2:1 mol ratio.

We will calculate the moles of barium hydroxide on dividing its grams by its molar mass.

Molar mass of Barium hydroxide is given as 171.3 g per mol.

$1.00gBa(OH)_2((1mol)/(171.3g))$

= 0.00584 mol $Ba(OH)_2$

Using mol ratio we calculate the moles of HCl as:

$0.00584molBa(OH)_2((2molHCl)/(1molBa(OH)_2))$

= 0.01168 mol HCl

We know that molarity is moles of solute per liter of solution. We have 0.01168 moles of HCl and its molarity is 0.100 M. So, we can calculate the liters of HCl solution used on dividing the moles by molarity as and on multiplying by 1000 the liters are converted to mL since, 1 L = 1000 mL.

$0.01168mol((1L)/(0.100mol))((1000mL)/(1L))$

= 116.8 mL

It could be round to 117 mL.

So, 117 mL of HCl are required.

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A particular refrigerant cools by evaporating liquefied CCl 2F 2. How many kg of the liquid must be evaporated to freeze a tray of water to ice (at zero degrees C)? The tray contains 525 grams water. Molar heat of fusion of ice = 6.01 kJ/mol. Molar heat of vaporization of CCl 2F 2 = 17.4 kJ/mole

Answers

Answer : The mass of $CCl_2F_2$ evaporated must be, 1.217 kg

Explanation :

First we have to calculate the moles of water.

$\text{Moles of water}=\frac{\text{Mass of water}}{\text{Molar mass of water}}$

Molar mass of water = 18 g/mol

$\text{Moles of water}=(525g)/(18g/mol)=29.17mol$

Now we have to calculate the heat released.

Heat released = Moles of water × Molar heat of fusion of ice

Heat released = 29.17 mol × 6.01 kJ/mol

Heat released = 175.3 kJ

Now we have to calculate the moles of $CCl_2F_2$

Heat = Moles of $CCl_2F_2$ × Molar heat of vaporization of $CCl_2F_2$

175.3 kJ = Moles of $CCl_2F_2$ × 17.4 kJ/mol

Moles of $CCl_2F_2$ = 10.07 mol

Now we have to calculate the mass of $CCl_2F_2$

$\text{Mass of }CCl_2F_2=\text{Moles of }CCl_2F_2* \text{Molar mass of }CCl_2F_2$

Molar mass of $CCl_2F_2$ = 120.9 g/mol

$\text{Mass of }CCl_2F_2=10.07mol* 120.9g/mol=1217.463g=1.217kg$

Thus, the mass of $CCl_2F_2$ evaporated must be, 1.217 kg

A chemist titrates 80.0mL of a 0.3184M pyridine C5H5N solution with 0.5397M HBr solution at 25°C . Calculate the pH at equivalence. The pKb of pyridine is 8.77.

Answers

Answer:pH = 2.96

Explanation:

C5H5N + HBr --------------> C5H5N+ + Br-

millimoles of pyridine = 80 x 0.3184 =25.472mM

25.472 millimoles of HBr must be added to reach equivalence point.

25.472 = V x 0.5397

V =25.472/0.5397= 47.197 mL HBr

total volume = 80 + 47.197= 127.196 mL

Concentration of [C5H5N+] = no of moles / volume=

25.472/ 127.196= 0.20M

so,

pOH = 1/2 [pKw + pKa + log C]

pKb = 8.77

pOH = 1/2 [14 + 8.77 + log 0.20]

pOH = 11.0355

pH = 14 - 11.0355

pH = 2.96

Magnesium (used in the manufacture of light alloys) reacts with iron(III) chloride to form magnesium chloride and iron. A mixture of 41.0 g of magnesium and 175.0 g of iron(III) chloride is allowed to react. Identify the limiting reactant and determine the mass of the excess reactant present in the vessel when the reaction is complete.

Answers

Answer: The limiting reactant is magnesium and mass of excess reactant present in the vessel is 96.35 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For magnesium:

Given mass of magnesium = 41.0 g

Molar mass of magnesium = 24 g/mol

Putting values in equation 1, we get:

$\text{Moles of magnesium}=(41.0g)/(24g/mol)=1.708mol$

For iron(III) chloride:

Given mass of iron(III) chloride = 175.0 g

Molar mass of iron(III) chloride = 162.2 g/mol

Putting values in equation 1, we get:

$\text{Moles of iron(III) chloride}=(175g)/(162.2g/mol)=1.708mol$

The chemical equation for the reaction of magnesium and iron(III) chloride follows:

$3Mg+2FeCl_3\rightarrow 3MgCl_2+2Fe$

By Stoichiometry of the reaction:

3 moles of magnesium reacts with 2 moles of iron(III) chloride

So, 1.708 moles of magnesium will react with = $(2)/(3)* 1.708=1.114mol$ of iron(III) chloride

As, given amount of iron(III) chloride is more than the required amount. So, it is considered as an excess reagent.

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

Moles of excess reactant left (iron(III) chloride) = [1.708 - 1.114] = 0.594 moles

Now, calculating the mass of iron(III) chloride from equation 1, we get:

Molar mass of iron(III) chloride = 162.2 g/mol

Moles of iron(III) chloride = 0.594 moles

Putting values in equation 1, we get:

$0.594mol=\frac{\text{Mass of iron(III) chloride}}{162.2g/mol}\n\n\text{Mass of iron(III) chloride}=(0.594mol* 162.2g/mol)=96.35g$

Hence, the limiting reactant is magnesium and mass of excess reactant present in the vessel is 96.35 grams.

Which step would help a student find the molecular formula of a compound from the empirical formula? Multiply the subscripts of the empirical formula by the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound. Subtract the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound from the subscripts of the empirical formula. Divide the subscripts of the empirical formula by the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound. Add the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound to the subscripts of the empirical formula..

Answers

Answer:

Multiply the subscripts of the empirical formula by the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound.

Explanation:

got it right on edge 2020 :)

Answer:

Multiply the subscripts of the empirical formula by the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound.

Explanation:

Agas has a pressure of 1.21 atm and a volume of 1.04
L. What is the new volume at 0.671 atm?

Answers

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Construction crews sometimes use this reaction for welding underwater structures:Fe2O3 + 2Al → Al2O3 + 2Fe

How many moles of iron (Fe) would be produced if 2.50 mol Fe2O3 react? Make sure to use the correct number of significant figures in your answer.

2.50 mol Fe2O3 =