Ask question

Ask question

All categories

2 years ago

9

6.74 g of the monoprotic acid KHP (MW = 204.2 g/mol) is dissolved into water. The sample is titrated with a 0.703 M solution of

calcium hydroxide to the equivalence point. What volume of base was used?

1 answer:

Dennis_Churaev [7]2 years ago

4 0

Answer:

Volume of the calcium hydroxide solution used is 0.0235 mL.

Explanation:

$2KHP+Ca(OH)_2\rightarrow 2H_2O+Ca(KP)_2$

Moles of KHP = $\frac{6.74 g}{204.2 g/mol}=0.0330 mol$

According to reaction, 2 moles of KHP with 1 mole of calcium hydroxide , then 0.0330 moles of KHP will recat with ;

$\frac{1}{2}\times 0.0330 mol=0.01650 mol$ of calcium hydroxide

Molarity of the calcium hydroxide solution = 0.703 M

Volume of calcium hydroxide solution = V

$Molarity=\frac{Moles}{Volume(L)}$

$0.703 M=\frac{0.01650 M}{V}$

$V=\frac{0.01650 M}{0.703 M}=0.0235 mL$

Volume of the calcium hydroxide solution used is 0.0235 mL.

You might be interested in

The Mond process produces pure nickel metal via the thermal decomposition of nickel tetracarbonyl: Ni(CO)4 (l) → Ni (s) + 4CO (g

Yuki888 [10]

<u>Answer:</u> The volume of CO formed is 254.43 L.

<u>Explanation:</u>

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

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

Given mass of $Ni(CO)_4$ = 444 g

Molar mass of $Ni(CO)_4$ = 170.73 g/mol

Putting values in above equation, we get:

$\text{Moles of }Ni(CO)_4=\frac{444g}{170.73g/mol}=2.60mol$

For the given chemical reaction:

$Ni(CO)_4(l)\rightarrow Ni(s)+4CO(g)$

By stoichiometry of the reaction:

1 mole of nickel tetracarbonyl produces 4 moles of carbon monoxide.

So, 2.60 moles of nickel tetracarbonyl will produce = $\frac{4}{1}\times 2.60=10.4mol$ of carbon monoxide.

Now, to calculate the volume of the gas, we use ideal gas equation, which is:

PV = nRT

where,

P = Pressure of the gas = 752 torr

V = Volume of the gas = ? L

n = Number of moles of gas = 10.4 mol

R = Gas constant = $62.364\text{ L Torr }mol^{-1}K^{-1}$

T = Temperature of the gas = $22^oC=(273+22)K=295K$

Putting values in above equation, we get:

$752torr\times V=10.4mol\times 62.364\text{ L Torr }mol^{-1}K^{-1}\times 295K\\\\V=254.43L$

Hence, the volume of CO formed is 254.43 L.

5 0

2 years ago

True or false: In a combustion reaction, matter is destroyed.

sp2606 [1]

Matter is never destroyed, nor created. So, this is false.

6 0

1 year ago

Read 2 more answers

Based on the results of your titration, what volume of the sample would an adult need to consume to reach the recommended daily

Mekhanik [1.2K]

Answer: 85.7 mL

Explanation:

Given the information from the question as plotted in the graph i will be uploading along side this answer,

Average of total volume of DCPIP used is

= (1.21 + 1.11 + 1.06)mL / 3

= 1.12 mL

and corresponding ( ascorbic acid ) is 0.70 g/L

Two parameter given as volume of DCPIP in final syringe and total volume of DCPIP are quite ambiguous

700mg ⇒ 1 L

THEREFORE volume that contains 60mg = (1000/700) × 60 = 85.7 mL

5 0

2 years ago

A 92.8 gram sample of CuSO4•5H2O.(copper sulfate pentahydrate) is heated until water is released. How many grams of water were r

Ivan

Mass of water released =

$92.8 g CuSO_{4}.5H_{2}O$ × $\frac{5 * 18 g H_{2}O}{249.68 g CuSO_{4}.5H_{2}O}$

= 33.45 g $H_{2}O$

7 0

1 year ago

Read 2 more answers

A) How many moles of CO2 and H2O are formed from 3.85 mole of propane C3H8 (This calculation needs to be done twice-once fro CO2

4vir4ik [10]

Answer:

a) 11.55 moles of carbon dioxide gas are formed from 3.85 mole of propane.

15.4 moles of water are formed from 3.85 mole of propane.

b)0.5176 moles of water are formed from 0.647 mole of oxygen gas.

0.1294 moles of propane are consumed.

Explanation:

$C_3H_7+5O_2\rightarrow 3CO_2+4H_2O$

a) Moles of propane = 3.85 moles

According to reaction, 1 mole of propane gives 3 moles of carbon dioxide gas.

Then 3.85 moles of propane will give:

$\frac{3}{1}\times 3.85 mol=11.55 mol$ of carbon dioxide gas.

11.55 moles of carbon dioxide gas are formed from 3.85 mole of propane.

According to reaction, 1 mole of propane gives 4 moles of water gas.

Then 3.85 moles of propane will give:

$\frac{4}{1}\times 3.85 mol=15.4 mol$ of water .

15.4 moles of water are formed from 3.85 mole of propane.

b) Moles of oxygen gas = 0.647 moles

According to reaction, 5 mole of oxygen gas gives 4 moles of water.

Then 0.647 moles of oxygen will give:

$\frac{4}{5}\times 0.647 mol=0.5176 mol$ of water.

0.5176 moles of water are formed from 0.647 mole of oxygen gas.

According to reaction, 5 mole of oxygen gas reacts with 1 mole of propane.

Then 0.647 moles of oxygen will give:

$\frac{1}{5}\times 0.647 mol=0.1294 mol$ of propane.

0.1294 moles of propane are consumed.

5 0

1 year ago