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3 years ago

How many grams of NH3 can be dissolved in 50 grams of water at 50oC?

Chemistry

1 answer:

weqwewe [10]3 years ago

5 0

15 grams of NH3 can be dissolved

<h3>Further explanation</h3>

Given

50 grams of water at 50°C

Required

mass of NH3

Solution

Solubility is the maximum amount of a substance that can dissolve in some solvents. Factors that affect solubility

1. Temperature:
2. Surface area:
3. Solvent type:
4. Stirring process:

We can use solubility chart (attached) to determine the solubility of NH3 at 50°C

From the graph, we can see that the solubility of NH3 in 100 g of water at 50 C is 30 g

So that the solubility in 50 grams of water is:

= 50/100 x 30

= 15 grams

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How many grams of S are in 475 g of SO2?

enot [183]

There are 237. 5 g of Sulfur,S in 475 g of SO2?

<h3 />

<h3>Calculation of grams of Sulfur</h3>

From the question, we can say that

The molar mass of sulfur = 32 g/mol

The molar mass of oxygen = 16 g/mol

Therefore,

The molar mass for SO2 = 32 + (16 × 2) g/mol = 64 g/mol

Now,

If 1 mole of SO2 contains 1 mole of S

Then 64 g of SO2, will contain 32g of S;

Such that

475 g of SO2 will give { $\frac{ 32g of S) ( 475 g of SO2)}{64 of SO2}$ } = 237. 5 g of Sulfur.

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2 years ago

I NEED HELP PLEASE, THANKS! :)

krok68 [10]

Answer:

$\large \boxed{\text{2.20 g Pb}}$

Explanation:

They gave us the masses of two reactants and asked us to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 239.27 32.00 207.2

2PbS + 3O₂ ⟶ 2Pb + 2SO₃

m/g: 2.54 1.88

2. Calculate the moles of each reactant

$\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}$

3. Calculate the moles of Pb from each reactant

$\textbf{From PbS:}\\\text{Moles of Pb} = \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}$

4. Calculate the mass of Pb

$\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}$

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3 years ago