Magnesium metal a gray solid, is heated in a crucble in the presence of oxygen

Using standard heats of formation, calculate the standard enthalpy change for the following reaction.

Katena32 [7]

Answer:

ΔH°r = -483.64 kJ

Explanation:

Let's consider the following balanced equation.

2 H₂(g) + O₂(g) ⇒ 2 H₂O(g)

We can calculate the standard enthalpy change of the reaction (ΔH°r) using the following expression.

ΔH°r = ∑ΔH°f(p) × np - ∑ΔH°f(r) × nr

where

ΔH°f: standard heat of formation

n: moles

p: products

r: reactants

ΔH°r = ΔH°f(H₂O(g)) × 2 mol - ΔH°f(H₂(g)) × 2 mol - ΔH°f(O₂(g)) × 1 mol

ΔH°r = (-241.82 kJ/mol) × 2 mol - 0 kJ/mol × 2 mol - 0 kJ/mol × 1 mol

ΔH°r = -483.64 kJ

4 0

3 years ago

How many grams of sodium oxide can be produced when 55.3 g Na react with 64.3 g O2?. . Unbalanced equation: Na + O2 → Na2O. . Sh

GenaCL600 [577]

this is a limiting reagent problem.

first, balance the equation
4Na+ O2 ---> 2Na2O

use both the mass of Na and mass of O2 to figure out how much possible Na2O you could make.
start with Na and go to grams of Na2O

55.3 gNa x (1molNa/23.0gNa) x (2 molNa2O/4 molNa) x (62.0gNa2O/1molNa2O) = 75.5 gNa2O

do the same with O2

64.3 gO2 x (1 molO2/32.0gO2) x (2 molNa2O/1 mol O2) x (62.0gNa2O/1molNa2O) = 249.2 g Na2O

now you must pick the least amount of Na2O for the one that you actually get in the reaction. This is because you have to have both reacts still present for a reaction to occur. So after the Na runs out when it makes 75.5 gNa2O with O2, the reaction stops.

So, the mass of sodium oxide is

75.5 g

3 0

3 years ago

Read 2 more answers

What is the volume of SO3 in 835 g SO3?

Pie

233.856 , sorry if i’m wrong

7 0

3 years ago

Sulfur dioxide, SO 2 ( g ) , can react with oxygen to produce sulfur trioxide, SO 3 ( g ) , by the reaction 2 SO 2 ( g ) + O 2 (

aleksley [76]

Answer: The amount of heat produced by the reaction is -21.36 kJ

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]$

For the given chemical reaction:

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2\times \Delta H_f_{(SO_3(g))})]-[(2\times \Delta H_f_{(SO_2(g))})+(1\times \Delta H_f_{(O_2(g))})]$

We are given:

$\Delta H_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H_f_{(SO_3(g))}=-395.7kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(2\times (-395.7))]-[(2\times (-296.8))+(1\times (0))]\\\\\Delta H_{rxn}=-197.8kJ/mol$

To calculate the number of moles, we use ideal gas equation, which is:

$PV=nRT$

where,

P = pressure of the gas = 1.00 bar

V = Volume of the gas = 2.67 L

n = number of moles of gas = ?

R = Gas constant = $0.0831\text{ L. bar }mol^{-1}K^{-1}$

T = temperature of the mixture = $25^oC=[25+273]K=298K$

Putting values in above equation, we get:

$1.00bar\times 2.67L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{1\times 2.67}{0.0831\times 298}=0.108mol$

To calculate the heat released of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat released = ?

n = number of moles = 0.108 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction = -197.8 kJ/mol

Putting values in above equation, we get:

$-197.8kJ/mol=\frac{q}{0.108mol}\\\\q=(-197.8kJ/mol\times 0.108mol)=-21.36kJ$

Hence, the amount of heat produced by the reaction is -21.36 kJ

3 0

3 years ago

The density of silver is 10.5 g/cm3. What is the mass (in kilograms) of a cube of silver that measures 0.94 m on each side?

Rzqust [24]

Answer:

8721.132 Kg.

Explanation:

The following data were obtained from the question:

Density (D) of silver = 10.5 g/cm³

Length (L) of silver = 0.94 m

Mass (m) of silver =.?

Next, we shall determine the volume of silver. This can be obtained as follow:

Volume = Length × Length × Length

Volume = L × L × L

Volume = L³

Length (L) of silver = 0.94 m

Volume (V) of silver =?

Volume (V) of silver = 0.94³

Volume (V) of silver = 0.830584 m³

Next, we shall convert 0.830584 m³ to cm³. This can be obtained as follow:

1 m³ = 1×10⁶ cm³

Therefore,

0.830584 m³ = 0.830584 m³ / 1 m³ × 1×10⁶ cm³

0.830584 m³ = 830584 cm³

Therefore, 0.830584 m³ is equivalent 830584 cm³.

Thus, the volume of the silver is 830584 cm³.

Next, we shall determine the mass of the silver. This can be obtained as follow:

Density (D) of silver = 10.5 g/cm³

Volume of the silver = 830584 cm³.

Mass of silver =.?

Density = mass /

10.5 = mass of silver /830584

Cross multiply

Mass of silver = 10.5 × 830584

Mass of silver = 8721132 g

Finally, we shall convert the mass of silver, 8721132 g to kilogram (kg). This can be obtained as follow:

1000 g = 1 Kg

Therefore,

8721132 g = 8721132 g / 1000 g × 1 Kg

8721132 g = 8721.132 Kg

Therefore, the mass of each cube of silver is 8721.132 Kg

3 0

3 years ago