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lorasvet [3.4K]
3 years ago
14

How many grams of Na2So4 would be formed if 0.75 moles of NaOH reacted?

Chemistry
1 answer:
Nata [24]3 years ago
4 0

Answer:

\boxed{\text{53 g }}

Explanation:

You don't give the reaction, but we can get by just by balancing atoms of Na.

We know we will need the partially balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.

M_r:                                 142.04  

             2NaOH + … ⟶ Na₂SO₄ + …  

n/mol:      0.75

1. Use the molar ratio of Na₂SO₄ to NaOH to calculate the moles of NaF.

Moles of Na₂SO₄ = 0.75 mol NaOH × (1 mol Na₂SO₄/2 mol NaOH

= 0.375 mol Na₂SO₄

2. Use the molar mass of Na₂SO₄ to calculate the mass of Na₂SO₄.

Mass of Na₂SO₄ = 0.375 mol Na₂SO₄ × (142.04 g Na₂SO₄/1 mol Na₂SO₄) = 53 g Na₂SO₄

The reaction produces \boxed{\text{53 g }} of Na₂SO₄.

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If you mix a 25.0mL sample of a 1.20m potassium sulfide solution with 15.0 mL of a 0.900m basium nitrate solution and the reacti
ddd [48]

Answer:

Limiting reagent: barium nitrate

Theoretical yield: 2.29 g BaS

Percent yield: 87%

Explanation:

The corrected balanced reaction equation is:

K₂S + Ba(NO₃)₂ ⇒ 2KNO₃ + BaS

The amount of potassium sulfide added is:

(25.0 mL)(1.20mol/L) = 30 mmol

The amount of barium nitrate added is:

(15.0mL)(0.900mol/L = 13.5 mmol

Since the reactant react in a 1:1 molar ratio, barium nitrate is the limiting reagent. If all of the limiting reagent reacts, the amount of barium sulfide produced is:

(13.5 mmol Ba(NO₃)₂)(BaS/Ba(NO₃)₂ ) = 13.5 mmol BaS

Converting this amount to grams gives the theoretical yield of BaS (molar mass 169.39 g/mol).

(13.5 mmol)(169.39 g/mol)(1g/1000mg) = 2.29 g BaS

The percent yield is calculated as follows:

(actual yield) / (theoretical yield) x 100%

(2.0 g) / (2.29 g) x 100% = 87%

6 0
3 years ago
4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)
Crazy boy [7]

1. The volume of ammonia consumed in the reaction is 23.2 L

2. The volume of oxygen consumed in the reaction is 29 L

<h3>Balanced equation</h3>

4NH₃(g) + 5O₂(g) --> 4NO(g) + 6H₂O(g)

From the balanced equation above,

6 L of H₂O were produced from 4 L of NH₃ and 5 L of O₂

<h3>1. How to determine the volume of ammonia, NH₃ consumed</h3>

From the balanced equation above,

6 L of H₂O were produced from 4 L of NH₃

Therefore,

34.8 L of H₂O will be produced from = (34.8 × 4) / 6 = 23.2 L of NH₃

Thus, 23.2 L of NH₃ were consumed

<h3>2. How to determine the volume of oxygen, O₂ consumed</h3>

From the balanced equation above,

6 L of H₂O were produced from 5 L of O₂

Therefore,

34.8 L of H₂O will be produced from = (34.8 × 5) / 6 = 29 L of O₂

Thus, 29 L of O₂ were consumed

Learn more about stoichiometry:

brainly.com/question/14735801

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7 0
1 year ago
What is a factor that affects the outcome of an experiment?
Anni [7]

Answer:

Observation affects the outcome

7 0
3 years ago
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Which of the following is the correct definition of an inherited behavior?
emmainna [20.7K]

Answer:

behaviors that are passed down genetically

4 0
3 years ago
20 ml of a mixture of C2H6 and C3H8 gases in X : Y mole ratio, is burnt completely by which 52 ml CO2 gas is evolved. If 20 ml o
Alja [10]

The volume of CO2 prodiced when 12 mL of CH4 and 8 mL of C4H10 is burnt is 56 mL.

<h3>What is the equation of the reaction?</h3>

The equation of the combustion of C2H6, CH4, C4H10, and C3H8 gases are given below:

2 C2H6 + 7 O2 ---> 4 CO2 + 6 H20

C3H8 + 5 O2 ---> 3 CO2 + 4 H20

CH4 + 2 O2 ----> CO2 + 2 H2O

2 C4H10 + 13 O2 ---> 8 CO2 + 10 H2O

X moles of C2H6 will produce 2X moles of CO2

Y moles of C3H8 will produce 3Y moles of CO2

Thus;

X + Y = 20 mL

X =20 - Y ---(1)

2X + 3Y = 52 mL ---(2)

Solving for X by substituting in (2)

2 (20 - Y) + 3Y = 52

40 - 2Y + 3Y = 52

Y = 12 mL

Then X = 8 mL

Volume of CO2 produced when 12 mL of CH4 and 8 mL of C4H10 is burnt = (12 × 2 + 8 × 4) = 56 mL

Therefore, the volume of CO2 produced is 56 mL.

Learn more about moles of gas and volume at: brainly.com/question/23883946

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4 0
2 years ago
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