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abruzzese [7]
2 years ago
8

3. Given 20g of Barium Hydroxide, how many grams of

Chemistry
1 answer:
anastassius [24]2 years ago
6 0

The number of grams of ammonium nitrate (NH₄NO₃) it would take for all the barium hydroxide to react is 18.7g

First, we will write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

Ba(OH)₂ + 2NH₄NO₃ → 2NH₄OH + Ba(NO₃)₂

This means, 1 mole of barium hydroxide is required to react with 2 moles of ammonium nitrate

Now, we will calculate the number of moles of barium hydroxide present.

Mass of barium hydroxide (Ba(OH)₂) = 20 g

Using the formula

Number\ of\ moles = \frac{Mass}{Molar\ mass}

Molar mass of Ba(OH)₂ = 171.34 g/mol

∴ Number of moles of Ba(OH)₂ present =\frac{20}{171.34}

Number of moles of Ba(OH)₂ present = 0.116727 mole

Now,

Since 1 mole of barium hydroxide is required to react with 2 moles of ammonium nitrate

Then,

0.116727 mole of barium hydroxide will react with 2 × 0.116727 mole of ammonium nitrate

2 × 0.116727 = 0.233454 mole

∴ Number of moles of NH₄NO₃ required is 0.233454 mole

Now, for the mass of ammonium nitrate (NH₄NO₃) required

From the formula

Mass = Number of moles × Molar mass

Molar mass of NH₄NO₃ = 80.043 g/mol

∴ Mass of NH₄NO₃ required = 0.233454 × 80.043

Mass of NH₄NO₃ required = 18.68636 g

Mass of NH₄NO₃ required ≅ 18.7g

Hence, the number of grams of ammonium nitrate (NH₄NO₃) it would take for all the barium hydroxide to react is 18.7g

Learn more on determining mass of reactant required here: brainly.com/question/11232389

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50.0 mL of a solution of HCl is combined with 100.0 mL of 1.05M NaOH in a calorimeter. The reaction mixture is initially at 22.4
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Answer:

2.1 M is the molarity of the HCl solution.

Explanation:

HCl+NaOH\rightarrow H_2O+NaCl

Molarity of HCl solution = M_1=?

Volume of HCl solution = V_1=50.0mL

Ionizable hydrogen ions in HCl = n_1=1

Molarity of NaOH solution = M_2=1.05 M

Volume of NaOH solution = V_2=100.0 mL

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n_1M_1V_1=n_2M_2V_2 (neutralization )

M_1=\frac{M_2V_2}{V_1}=\frac{1.05M\times 100.0 mL}{50.0 mL}

M_1=2.1 M

2.1 M is the molarity of the HCl solution.

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3 years ago
Determine the number of moles of air present in 1.35 L at 750 torr and 17.0°C.
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Your answers are correct. I love the Universal formula.
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A solid form of matter in which there is long range repeating order is called ________
polet [3.4K]

Answer: A solid form of matter in which there is long-range repeating order is called crystalline.

Explanation:

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Gaseous ethane (CH,CH,) will react with gaseous oxygen (0,) to produce gaseous carbon dioxide (CO) and gaseous water (H2O). Supp
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Answer:

0.00 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.  

Mᵣ            30.07      32.00  

              2CH₃CH₃ + 7O₂ ⟶ 4CO₂ + 6H₂O

Mass/g:      1.50          11.

2. Calculate the moles of each reactant  

\text{moles of C$_{2}$H}_{6} = \text{1.50 g C$_{2}$H}_{6} \times \dfrac{\text{1 mol C$_{2}$H}_{6}}{\text{30.07 g C$_{2}$H}_{6}} = \text{0.04988 mol C$_{2}$H}_{6}\\\\\text{moles of O}_{2} = \text{11. g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.34 mol O}_{2}

3. Calculate the moles of CO₂ we can obtain from each reactant

From ethane:

The molar ratio is 4 mol CO₂:2 mol C₂H₆

\text{Moles of CO}_{2} = \text{0.04988 mol C$_{2}$H}_{6} \times \dfrac{\text{4 mol CO}_{2}}{\text{2 mol C$_{2}$H}_{6}} = \text{0.09976 mol CO}_{2}

From oxygen:

The molar ratio is 4 mol CO₂:7 mol O₂

\text{Moles of CO}_{2} =  \text{0.34 mol O}_{2}\times \dfrac{\text{4 mol CO}_{2}}{\text{7 mol O}_{2}} = \text{0.20 mol CO}_{2}

4. Identify the limiting and excess reactants

The limiting reactant is ethane, because it gives the smaller amount of CO₂.

The excess reactant is oxygen.

5. Mass of ethane left over.

Ethane is the limiting reactant. It will be completely used up.

The mass of ethane left over will be 0.00 g.

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Answer:

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