Which of these are the largest? asteroids, comets, meteors, or are they all the same size?

Calculate the grams of sulfur dioxide, SO2, produced when a mixture of 35.0 g of carbon disulfide and 30.0 g of oxygen reacts. W

alekssr [168]

Answer:

58.9g of SO2 is produced

8g of oxygen remains unconsumed

Explanation:

The burning of Carbon disulfide (CS2) in oxygen. gives the reaction:

CS2 (g) + 3O2 (g) → CO2 (g) + 2SO2 (g)

Molar mass of CS2 = 76.139 g/mol

Molar mass of O2 = 15.99 g/mol

Molar mass of SO2 = 64.066 g/mol

Number of moles of CS2 = 35g/ 76.139 g/mol =0.46 moles

Number of moles of O2 = 30g/15.999 g/mol =1.88 moles

From the chemical reaction

1 mole of CS2 react with 3 moles of O2 to give 2 moles of SO2

Thus 0.46 moles of CS2 reacts to form 2× 0.46 = 0.92 moles of SO2

Mass of SO2 produced = 0.92×64.07 = 58.9g of SO2 is produced

thus 0.46 moles of CS2 reacts with 3 × 0.46 moles of O2 which is =1.38 moles of O2

Thus oxygen is the limiting reactant with 1.88 - 1.38 = 0.496~~0.5 mole remaining

Or 8g of oxygen

58.9g of SO2 is produced

oxygen is the limiting

4 0

3 years ago

Calculate the concentration of hi when the equilibrium constant is 1x10^5

exis [7]

Answer is: concentration of hydrogen iodide is 6 M.

Balanced chemical reaction: H₂(g) + I₂(g) ⇄ 2HI(g).
[H₂] = 0.04 M; equilibrium concentration of hydrogen.
[I₂] = 0.009 M; equilibrium concentration of iodine.
Keq = 1·10⁵.
Keq = [HI]² / [H₂]·[I₂].
[HI]² = [H₂]·[I₂]·Keq.
[HI]² = 0.04 M · 0.009 M · 1·10⁵.
[HI]² = 36 M².
[HI] = √36 M².
[HI] = 6 M.

5 0

3 years ago

Propane reacts with 3.15moles of oxygen how much propane reacts with oxygen

Inessa [10]

27.7g

Explanation:

Given parameters:

Number of moles of oxygen = 3.15moles

Unknown:

Mass of reacting propane = ?

Solution:

Equation of the reaction:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

We work from the known to the unknown parameters.

The known in this expression is the number of moles of oxygen.

In the reaction:

5 moles of oxygen gas reacts with 1 mole of propane

3.15 mole of oxgen gas will react with $\frac{3.15}{5}$ = 0.63moles

Number of moles of reacting propane gas is 0.63moles

To find the mass;

mass = number of moles x molar mass = 0.63 x molar mass

molar mass of propane = (3 x 12) + (8x1) = 44g/mol

mass of propane = 0.63 x 44 = 27.7g

Learn more:

Combustion reaction brainly.com/question/6126420

#learnwithBrainly

8 0

3 years ago

An aqueous solution containing 9.82 g9.82 g of lead(II) nitrate is added to an aqueous solution containing 5.76 g5.76 g of potas

n200080 [17]

Answer:

The limiting reactant is lead(II) nitrate.

7.20 g of precipitate are formed.

1.9 g of the excess reactant remain.

Explanation:

The reaction that takes place is:

Pb(NO₃)₂(aq) + 2KCl(aq) → PbCl₂(s) + 2KNO₃(aq)

With a percent yield of 87.5%.

To determine the limiting reactant, first we convert the masses of each reactant to moles, using their molar mass:

9.82 g Pb(NO₃)₂ ÷ 331.2 g/mol = 0.0296 mol Pb(NO₃)₂

5.76 g KCl ÷ 74.55 g/mol = 0.0773 mol KCl

Looking at the stoichiometric coefficients, we see that 1 mol of Pb(NO₃)₂ would react completely with 2 moles of KCl. Following that logic, 0.0296 mol Pb(NO₃)₂ would react completely with (2x0.0296) 0.0592 mol of KCl. We have more than that amount of KCl, this means KCl is the reactant in excess and Pb(NO₃)₂ is the limiting reactant.

To calculate the mass of precipitate (PbCl₂) formed, we use the moles of the limiting reactant:

0.0296 mol Pb(NO₃)₂ $\frac{1molPbCl_{2}}{1molPb(NO_{3})_{2}}$ * $\frac{278.1g}{1molPbCl_{2}}$ * 87.5/100 = 7.20 g PbCl₂

- Keeping in mind the reaction yield, the moles of Pb(NO₃)₂ that would react are:

0.0296 mol Pb(NO₃)₂ * 87.5/100 = 0.0259 mol Pb(NO₃)₂

Now we convert that amount to moles of KCl and finally into grams of KCl:

0.0259 mol Pb(NO₃)₂ $\frac{2molKCl}{1molPb(NO_{3})_2}$ * $\frac{74.55g}{1molKCl}$ = 3.86 g KCl

3.86 g of KCl would react, so the amount remaining would be:

5.76 - 3.86 = 1.9 g KCl

8 0

3 years ago

Br2 + NaF —> Which best completes the chemical equation?

gizmo_the_mogwai [7]

When Br2 and NaF are mixed, no reaction would occur since Br cannot replace F in NaF which gives NaBr + F₂ as products. Hence, Fluorine is more reactive than Bromine. Hence, Br can't replace F.

Explanation:

When two reactants Br2 and NaF are mixed, no reaction takes place since Br cannot replace F in NaF. By seeing the periodic table, the positions of the halogens in the periodic table, bromine is located way below fluorine.

Bromine will not have sufficient energy to replace fluorine and so it will not have energy which is sufficient to join because of low reactivity. Among these fluorine represents the most reactive element among halogens.

Some of you may think it is a single replacement reaction that gives NaBr + F₂ as products. But, according to the halogen reactivity, it decreases from top to down of the group. F is placed above Br. Hence, Fluorine is more reactive than Bromine. Hence, Br can't replace F.

6 0

3 years ago