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3 years ago

10

Given the balanced equation representing a reaction:

1 answer:

lapo4ka [179]3 years ago

8 0

The answer is (3) 112 g. The question gives the balanced reaction and specific that 88 grams of CO2 has completely reacted and produces 200 grams of CaCO3. So the mass of CaO equals 200-88=112 g.

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1.) The process for converting ammonia to nitric acid involves the conversion of NH3 to

Firdavs [7]

Answer:

a) 1.39 g ; b) O₂ is limiting reactant, NH₃ is excess reactant; c) 0.7 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

MM: 17.03 32.00 30.01

4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

Mass/g: 1.5 1.85

2. Calculate the moles of each reactant

$\text{moles of NH}_{3} = \text{1.5 g NH}_{3} \times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}} = \text{0.0881 mol NH}_{3}\\\\\text{moles of O}_{2} = \text{1.85 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.057 81 mol O}_{2}$

3. Calculate the moles of NO we can obtain from each reactant

From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

$\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}$

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

$\text{Moles of NO} = \text{0.057 81 mol O}_{2}\times \dfrac{\text{4 mol NO}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NO}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO.

The excess reactant is NH₃.

5. Calculate the mass of NO formed

$\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}$

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

$\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}$

7. Calculate the mass of NH₃ reacted

$\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}$

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

8 0

3 years ago

The mass and volume of each box is given in the chart below. Calculate the density of each box. Density = Mass/ volume which box

stira [4]

Box C will have the greatest density.

All boxes have the same volume.

Explanation:

We calculate the density using the following formula:

density = mass / volume

density of Box A = 10 g / 20 cm³ = 0.5 g/cm³

density of Box B = 30 g / 20 cm³ = 1.5 g/cm³

density of Box C = 170 g / 20 cm³ = 8.5 g/cm³

Box C will have the greatest density.

All boxes have the same volume.

Learn more about:

density

brainly.com/question/952755

#learnwithBrainly

7 0

3 years ago

PLEASE HELP!! 40 POINTS

PSYCHO15rus [73]

Answer:

1) 6.524779402×10^(-17)

2)521.1g

3)113

Explanation:

5 0

3 years ago

Read 2 more answers

Which group tends not to form ions or react?

Anastaziya [24]

Group 8 elements. They are unreactive and stable

3 0

3 years ago

16. Chemists can synthesize new materials using

Alex787 [66]

Answer:

Yes Concurred,but where is the question

8 0

3 years ago