Look at the following equation.

If we mix 25 grans if sodium oxide with a large amount of potassium chloride, how many grams of sodium chloride should be produc

amid [387]

Answer:

47.2 g

Explanation:

Data given:

mass of Sodium oxide (Na₂O) = 25 grams

mass of potassium chloride (KCl) = excess

amount of sodium chloride (NaCl) = ?

Solution:

First we look for reaction of sodium oxide with potassium chloride.

Na₂O + 2 KCl -----------> 2 NaCl + K₂O

As we know that potassium chloride is in excess amount and sodium oxide is 25 g so it means sodium oxide is a limiting reactant and amount of sodium chloride depends on the amount of sodium oxide.

So, now we will look for mole mole ration of Na₂O to NaCl.

Na₂O + 2 KCl -----------> 2 NaCl + K₂O

1 mol 2 mol

from above equation we come to know that 1 mole of Na₂O gives 2 moles of NaCl.

Now convert moles to mass

As we Know

Molar mass of Na₂O = 2(23) + 16 = 62 g/mol

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

So

Na₂O + 2 KCl -----------> 2 NaCl + K₂O

1 mol (62 g/mol) 2 mol (58.5 g/mol)

62 g 117 g

So, it means that 62 g of Na₂O produce 117 g of NaCl then how many grams of NaCl will be produce by 25 g of Na₂O

Apply unity formula

62 g of Na₂O ≅ 117 g of NaCl

25 g of Na₂O ≅ X g of NaCl

Do cross multiplication

X g of NaCl = 117 g x 25 g / 62 g

X g of NaCl = 47.2 g

So,

25 g of Na₂O gives 47.2 grams of NaCl.

4 0

4 years ago

What volume of a 0.550 M solution of potassium hydroxide (KOH) can be made with 19.9 g of potassium hydroxide?

Natalija [7]

Answer:

0.645 L

Explanation:

To find the volume, you need to (1) convert grams to moles (using the molar mass) and then (2) calculate the volume (using the molarity ratio). The final answer should have 3 sig figs to match the sig figs of the given values.

(Step 1)

Molar Mass (KOH): 39.098 g/mol + 15.998 g/mol + 1.008 g/mol

Molar Mass (KOH): 56.104 g/mol

19.9 grams KOH 1 mole
-------------------------- x ----------------------- = 0.355 moles KOH
56.014 grams

(Step 2)

Molarity = moles / volume <----- Molarity ratio

0.550 M = 0.355 moles / volume <----- Insert values

(0.550 M) x volume = 0.355 moles <----- Multiply both sides by volume

volume = 0.645 L <----- Divide both sides by 0.550

6 0

1 year ago

Ik i already asked this but i need diff point of views

salantis [7]

Answer:

Is it prescribe to you?If so than yes if not then no need to

Explanation:

5 0

3 years ago

Read 2 more answers

trial 2, 2.68 g/cm/3 trial 3, 2.84g/cm/3. aluminum has a density of 2.70g/cm/3 calcute the percent error for each trial.

serg [7]

Answer:

trial 2: 0.74

trial 3: 5.19

I think but the equation for sloving percent error is:

(true value - determined value)/true value * 100

Explanation:

6 0

4 years ago

please hurry! If 3.87g of powdered aluminum oxide is placed in a container containing 5.67g of water, what is the limiting react

VikaD [51]

Answer:

Explanation:

Given parameters:

Mass of aluminium oxide = 3.87g

Mass of water = 5.67g

Unknown:

Limiting reactant = ?

Solution:

The limiting reactant is the reactant in short supply in a chemical reaction. We need to first write the chemical equation and convert the masses given to the number of moles.

Using the number of moles, we can ascertain the limiting reactants;

Al₂O₃ + 3H₂O → 2Al(OH)₃

Number of moles;

Number of moles = $\frac{mass}{molar mass}$

molar mass of Al₂O₃ = (2x27) + 3(16) = 102g/mole

number of moles = $\frac{3.87}{102}$ = 0.04mole

molar mass of H₂O = 2(1) + 16 = 18g/mole

number of moles = $\frac{5.67}{18}$ = 0.32mole

From the reaction equation;

1 mole of Al₂O₃ reacted with 3 moles of H₂O

0.04 mole of Al₂O₃ will react with 3 x 0.04 mole = 0.12 mole of H₂O

But we were given 0.32 mole of H₂O and this is in excess of amount required.

This shows that Al₂O₃ is the limiting reactant

6 0

3 years ago