Which scientist bombarded gold sheets with alpha particles to look at the atom?

Calculate the moles of SO₂ produced when 5 moles of FeS₂ reacts according to the equation: 4FeS₂+ 11O₂ → 2Fe₂O₃ + 8SO₂

Gala2k [10]

Answer:

10 moles of SO₂ are produced when 5 moles of FeS₂

Explanation:

Stoichiometry: it is the theoretical proportion in which the chemical species are combined in a chemical reaction. The stoichiometric equation of a chemical reaction relates molecules or number of moles of all the reagents and products that participate in the reaction.

In other words, stoichiometry establishes relationships between the molecules or elements that make up the reactants of a chemical equation with the products of said reaction. The relationships established are molar relationships (that is, moles) between the compounds or elements that make up the chemical equation.

The stoichiometric coefficients of a chemical reaction indicate the proportion in which said substances react.

Taking into account the above, you can apply the following rule of three: by stoichiometry if 4 moles of FeS₂ produce 8 moles of SO₂, then when reacting 5 moles of FeS₂ how many moles of SO₂ will they produce?

$moles of SO_{2} =\frac{5 moles of FeS_{2} *8moles of SO_{2}}{4 moles of FeS_{2}}$

moles of SO₂= 10

<u><em>10 moles of SO₂ are produced when 5 moles of FeS₂</em></u>

7 0

2 years ago

Paul placed an ice cube into a beaker on a hot plate. The ice cube changed from a solid to a liquid and then to a gas. During th

faltersainse [42]

One thing that does not change is the chemical composition of water, which is still H2O. And maybe mass, if all of the particles remain inside the beaker, which was never mentioned in the question so I am not sure.

6 0

2 years ago

Read 2 more answers

Please help me as soon as possible! Please help me as soon as possible!

Nastasia [14]

The Answer is D. Suspending a heavy weight with a strong chain.

6 0

3 years ago

Hydrogen sulfide burns form sulfur dioxide:

Helga [31]

Answer: 404.04 kJ.

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

moles of $H_2S$

$\text{Number of moles}=\frac{26.7g}{34.1g/mol}=0.78moles$

$2H_2S(g)+3O_2(g)\rightarrow 2SO_2(g)+2H_2O(g)$ $\Delta H=-1036kJ$

According to stoichiometry :

2 moles of $H_2S$ on burning produces = 1036 kJ

Thus 0.78 moles of $H_2S$ on burning produces = $\frac{1036kJ}{2}\times 0.78=404.04$

Thus the enthalpy change when burning 26.7 g of hydrogen sulfide is 404.04 kJ.

7 0

2 years ago

A 0.964 gram sample of a mixture of sodium formate and sodium chloride is analyzed by adding sulfuric acid. The equation for the

frosja888 [35]

Answer: 67.8 %.

Explanation:

Okay, let us delve right into the solution to the question;

The balanced chemical reaction is given by the equation (1) below;

2 HCOONa + H2SO4 ---------> 2 CO + 2 H2O + Na2SO4. ----------------------------------------------------------------------------(1).

From the balanced chemical reaction in equation (1) above we can see that; 2 moles of HCOONa reacts with one moles of tetraoxosulphate acid, H2SO4 to produce 2 moles of carbonmonoxide,CO; 2 moles of water, H2O and 1 mole of sodium tetraoxosulphate, Na2SO4.

The parameters given from the question are; total atmospheric pressure, P(t) = 752 torr, volume of CO= 242 mL = 0.242 Litres.

STEP ONE : find the carbon monoxide,CO pressure; P(CO).

Using the formula below;

P(t) = P(CO) + P(H2O). Hence;

P(CO) = P(t) - P(H2O). Note that P(H2O)= 19.8 torr.

==>P(CO)= 752 torr - 19.8 torr = 732.2 torr.

STEP TWO: calculate the number of moles of Carbonmonoxide,CO.

Using the formula below;

Number of moles= pressure(P) × volume(v) / gas constant(R) × temperature (T).

That is, n= PV/RT.

n= 732 torr × 0.242 Litres/ 62.4 × 295.15.

= 9.62 × 10^-3 mol of CO.

STEP THREE:

2 moles of HCOONa = 2 moles of CO.

=> 2 moles of HCOONa = 2 moles of CO/ 2 moles of CO = 1 mol( HCOONa/ CO).

Then, 9.62 × 10^-3 mol of CO × 1 mol( HCOONa/ CO).

==> 9.62 × 10^-3 mol HCOONa × molar masss of HCOONa(68 grams/mol)

= 0.654 grams.

Therefore, the percentage of sodium formate in the original mixture = 0.654 grams/ 0.964 gram × 100 = 67.8 %.

5 0

2 years ago