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2 years ago

If 3.45 grams of Mg are burned how many grams of MgO are produced

Chemistry

2 answers:

artcher [175]2 years ago

5 0

Answer:

Mass of MgO produced = 5.72 g

Explanation:

$2Mg + O_2 \rightarrow 2MgO$

Mass of Mg = 3.45 g

Molar mass of Mg = 24.305 g/mol

No. of mol of Mg = $\frac{Mass\ in\ g}{Molar\ mass}$

No. of mol of Mg = $\frac{3.45}{24.305}=0.142 mol$

From the reaction coefficient,

2 mol of Mg form 2 mol of MgO

So, 0.142 mol of Mg form 0.142 mol of MgO

Molar mass of MgO = 40.3044 g/mol

Mass of MgO in g = No. of mol × Molar mass

= 0.142 mol × 40.3044 g/mol

= 5.72 g

coldgirl [10]2 years ago

4 0

The number of grams of Mgo that are produced is calculated as follows
calculate the moles of Mg used

that is moles =mass/molar mass
= 3.45 g/ 24.305g/mol = 0.142 moles

write the equation for reaction
that is 2 Mg + O2 = 2 MgO

by use of mole ratio between Mg to Mgo which is 2:2 this implies that the moles of Mgo is also 0.142 moles

mass of Mgo= moles x molar mass of MgO
molar mass of Mgo = 24.305 + ( 3x15.999) = 72.302 g/mol

mass= 0.142 mol x 72.302 g/mol= 10.27 grams

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consideras util conocer las propiedades extensivas e intensivas de los insumos utilizados para la elaboración de producto ¿por q

Brums [2.3K]

Answer:

Explanation:

No.

Las propiedades físicas de los materiales y sistemas a menudo se pueden clasificar como intensivas o extensivas, según cómo cambia la propiedad cuando cambia el tamaño (o extensión) del sistema. Según la IUPAC, una cantidad intensiva es aquella cuya magnitud es independiente del tamaño del sistema, mientras que una cantidad extensiva es aquella cuya magnitud es aditiva para los subsistemas. Esto refleja las ideas matemáticas correspondientes de media y medida, respectivamente.

Una propiedad intensiva es una propiedad a granel, lo que significa que es una propiedad física local de un sistema que no depende del tamaño del sistema o de la cantidad de material en el sistema. Los ejemplos de propiedades intensivas incluyen temperatura, T; índice de refracción, n; densidad, ρ; y dureza de un objeto.

Por el contrario, propiedades extensivas como la masa, el volumen y la entropía de los sistemas son aditivas para los subsistemas porque aumentan y disminuyen a medida que crecen y se reducen, respectivamente.

Estas dos categorías no son exhaustivas, ya que algunas propiedades, físicas no son exclusivamente intensivas ni extensivas. Por ejemplo, la impedancia eléctrica de dos subsistemas es aditiva cuando, y solo cuando, se combinan en serie; mientras que si se combinan en paralelo, la impedancia resultante es menor que la de cualquiera de los subsistemas.

¡Espero haberte ayudado! :)

7 0

2 years ago

Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll

makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature $T_1$ = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

The truncated viral equation is expressed as:

$\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}$

where; B = - $152.5 \ cm^3 /mol$ C = -5800 $cm^6/mol^2$

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

$\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

$4.138*10^{-4} \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

Multiplying through with V² ; we have

$4.138*10^4 \ V ^3 = V^2 - 152.5 V - 5800 = 0$

$4.138*10^4 \ V ^3 - V^2 + 152.5 V + 5800 = 0$

V = 2250.06 cm³ mol⁻¹

Z = $\frac{PV}{RT}$

Z = $\frac{1800*2250.06}{8.314*10^3*523.15}$

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

$T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345$

$T__{\gamma}} = \frac{T}{T_c}$

$T__{\gamma}} = \frac{523.15}{647.1}$

$T__{\gamma}} = 0.808$

$P__{\gamma}} = \frac{P}{P_c}$

$P__{\gamma}} = \frac{1800}{22055}$

$P__{\gamma}} = 0.0816$

$B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}$

$B_o = 0.083 - \frac{0.422}{0.808^{1.6}}$

$B_o = 0.51$

$B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}$

$B_1 = -0.282$

The compressibility is calculated as:

$Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}$

$Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}$

Z = 0.9386

$V= \frac{ZRT}{P}$

$V= \frac{0.9386*8.314*10^3*523.15}{1800}$

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At $T_1 = 523.15 \ K \ and \ P = 1800 \ k Pa$

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V = 0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = $\frac{PV}{RT}$

Z = $\frac{1800*10^3 *0.1249}{729.77*523.15}$

Z = 0.588

3 0

2 years ago

A vessel contained N2, Ar, He, and Ne. The total pressure in the vessel was 987 torr. The partial pressures of nitrogen, argon,

xz_007 [3.2K]

Answer:

The partial pressure of neon in the vessel was 239 torr.

Explanation:

In all cases involving gas mixtures, the total gas pressure is related to the partial pressures, that is, the pressures of the individual gaseous components of the mixture. Put simply, the partial pressure of a gas is the pressure it exerts on a mixture of gases.

Dalton's law states that the total pressure of a mixture of gases is equal to the sum of the pressures that each gas would exert if it were alone. Then:

PT= P1 + P2 + P3 + P4…+ Pn

where n is the amount of gases present in the mixture.

In this case:

PT=PN₂ + PAr + PHe + PNe

where:

PT= 987 torr
PN₂= 44 torr
PAr= 486 torr
PHe= 218 torr
PNe= ?

Replacing:

987 torr= 44 torr + 486 torr + 218 torr + PNe

Solving:

987 torr= 748 torr + PNe

PNe= 987 torr - 748 torr

PNe= 239 torr

<u><em>The partial pressure of neon in the vessel was 239 torr.</em></u>

4 0

2 years ago

A piece of wire contains 1.52x10^22 atoms of copper. Calculate the miles of copper in the wire

melamori03 [73]

Moles of Copper =

$1.52$ × $10^{22}atoms$ × $\frac{1 mol }{6.022 * 10^{23}atoms}$

= 0.0252 mol Cu

6 0

2 years ago

An alkene with the molecular formula C8H16 undergoes ozonolysis to yield a mixture of (CH3)2C=O and (CH3)3CCHO. The alkene is:

aalyn [17]

Answer:

2,4,4-trimethyl-2-pentene yields mixture of $(CH_{3})_{2}C=O$ and $(CH_{3})_{3}CHO$

Explanation:

In ozonolysis (hydrolysis step involve a reducing agent such as Zn, $Me_{2}S$ etc.), a pi bond is broken to form ketone/aldehyde.

Ketone is formed from di-substituted side of double bond and aldehyde is formed from mono-substituted side of double bond.

Ozoznolysis involves two consecutive steps : (1) formation of ozonide, (2) hydrolysis of ozonide.

Hydrolysis can be done with/without using reducing agent. Carboxylic acid/carbon dioxide/ketone is produced when hydrolysis is done without using reducing agent.

Here, 2,4,4-trimethyl-2-pentene yields mixture of $(CH_{3})_{2}C=O$ and $(CH_{3})_{3}CHO$

Reaction steps are shown below.

8 0

2 years ago