Which of these elements is a halogen? potassium (K) calcium (Ca) vanadium (V) gallium (Ga) bromine (Br)

How many grams of Na2SO4 will be formed from 200.0g of NaOH

nikklg [1K]

The question is incomplete, the complete question is;

Using the following equation 2 NaOH(aq) + H2SO4(aq) → 2 H2O(aq) + Na2SO4(aq) how many grams of sodium sulfate will be formed if you start with 200 grams of sodium hydroxide and you have an excess of sulfuric acid

Answer:

355.1 g

Explanation:

The equation of the reaction is;

2 NaOH(aq) + H2SO4(aq) → 2 H2O(aq) + Na2SO4(aq)

We have been told that H2SO4 is in excess so NaOH is the limiting reactant. Therefore;

Number of moles in 200g of NaOH = 200g/40g/mol = 5 moles

So;

2 moles of NaOH yields 1 mole of Na2SO4

5 moles of NaOH will yield 5 * 1/2 = 2.5 moles of Na2SO4

Molar mass of Na2SO4 = 142.04 g/mol

Mass of Na2SO4= 2.5 moles * 142.04 g/mol = 355.1 g

8 0

3 years ago

an aqeous solution of oxalic acid h2c2o4 was prepared by dissolving a 0.5842g of solute in enough water to make a 100 ml solutio

vampirchik [111]

The question is incomplete, here is the complete question:

An aqeous solution of oxalic acid was prepared by dissolving a 0.5842 g of solute in enough water to make a 100 ml solution a 10 ml aliquot of this solution was then transferred to a volumetric flask and diluted to a final volume of 250 ml. How many grams of oxalic acid are in 100. mL of the final solution?

<u>Answer:</u> The mass of oxalic acid in final solution is 0.0234 grams

<u>Explanation:</u>

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

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ ......(1)

Given mass of oxalic acid = 0.5842 g

Molar mass of oxalic acid = 90 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Molarity of oxalic acid solution}=\frac{0.5842\times 1000}{90\times 100}\\\\\text{Molarity of oxalic acid solution}=0.0649M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated oxalic acid solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted oxalic acid solution

We are given:

$M_1=0.0649M\\V_1=10mL\\M_2=?M\\V_2=250.0mL$

Putting values in above equation, we get:

$0.0649\times 10=M_2\times 250.0\\\\M_2=\frac{0.0649\times 10}{250}=0.0026M$

Now, calculating the mass of glucose by using equation 1, we get:

Molarity of oxalic acid solution = 0.0026 M

Molar mass of oxalic acid = 90 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0026=\frac{\text{Mass of oxalic acid solution}\times 1000}{90\times 100}\\\\\text{Mass of oxalic acid solution}=\frac{0.0026\times 90\times 100}{1000}=0.0234g$

Hence, the mass of oxalic acid in final solution is 0.0234 grams

6 0

4 years ago

Solutions can be composed of:

Anika [276]

A solution may exist in any phase so your answer is D. any of the above

hope this helps :)

7 0

4 years ago

There are two alloys, one is 1 part silver and 5 parts copper and the other 3 parts silver and 1 part copper are mixed to form 3

borishaifa [10]

Lets make x equal the number times you use the 1/5 ratio alloy and y equal the number of times you use the 3/1 ratio alloy. You can make the equation (x+3y)-(5x+y)=0 and (x+3y)+(5x+y)=350
Then you can make the system of equation of:
-4x+2y=0 (this is y=2x)
6x+4y=350
You can make 6x+8x=350 by through substitution and solve of x.
14x=350
x=25 (which means y=50)

that means that the weight of the 1/5 ratio alloy is 150 pounds.
(25+(5x25))=150
the weight of the 3/1 ratio alloy is 200 pounds.
((50x3)+50)=200

I hope this helps.

3 0

4 years ago

consider the reaction between calcium oxide and carbon dioxide: cao ( s ) + co 2 ( g ) → caco 3 ( s ) a chemist allows 14.4 g of

lana [24]

Answer:

Explanation:

Given data:

Mass of calcium oxide = 14.4 g

Mass of carbon dioxide = 13.8 g

Actual yield of calcium carbonate = 19.4 g

Mass of calcium carbonate produced = ?

Limiting reactant = ?

Percent yield = ?

Chemical equation:

CaO + CO₂ → CaCO₃

Number of moles of CaO:

Number of moles of CaO = Mass /molar mass

Number of moles of CaO = 14.4 g / 56.1g/mol

Number of moles of CaO = 0.26 mol

Number of moles of CO₂:

Number of moles of CO₂= Mass /molar mass

Number of moles of CO₂ = 13.8 g / 44 g/mol

Number of moles of CO₂ = 0.31 mol

Now we will compare the moles of CaCO₃ with CO₂ and CaO.

CaO : CaCO₃

1 : 1

0.26 : 0.26

CO₂ : CaCO₃

1 : 1

0.31 : 0.31

The number of moles of CaCO₃ produced by CaO are less it will be limiting reactant.

Limiting reactant:

CaO

Theoretical yield:

Mass of CaCO₃ = moles × molar mass

Mass of CaCO₃ = 0.26 mol × 100 g/mol

Mass of CaCO₃ = 26 g

Percent yield:

Percent yield = Actual yield / theoretical yield × 100

Percent yield = 19.4 g/ 26 g× 100

Percent yield = 74.6 %

3 0

3 years ago