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nekit [7.7K]
3 years ago
7

Stoichiometry: How much magnesium oxide, in grams, is produced from the reaction of 2 grams of magnesium?

Chemistry
1 answer:
STatiana [176]3 years ago
6 0
1) Chemical equation:

Mg + O2 -> MgO

2) Balanced chemical equation:

2Mg + O2 -> 2MgO

3) Ratios

2 mol Mg : 1 mol O2 : 2 mol MgO

4) Convert 2 grams of Mg to number of moles

Molar mass of Mg: 24.3 g/mol

# of moles = grams / molar mass = 2g / 24.3 g/mol =  0.0823 mol

5) Use the theoretical ratio fo find the number of moles of MgO produced

0.0823 mol Mg *   2mol MgO / 2 mol Mg = 0.0823 mol MgO

6) Use the molar mass of MgO to obtain the quantity in grams

Molar mass of MgO = 24.3 g/mol + 16 g/mol = 40.3 g/mol

0.0823 mol MgO * 40.3 g/ mol = 3.32 g

Answer: 3.32 g
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Volgvan
First, we need to get the concentration of  [NaH2PO4]:
[NaH2PO4] =( mass / molar mass ) * volume L
when we have mass NaH2PO4 = 6.6 g & molar mass = 120g/mol & V = 0.355 L
So by substitution:
[NaH2PO4] = (6.6g / 120g/mol) * 0.355 L = 0.0195 M

then, we need to get the concentration of [Na2HPO4]:
[Na2HPO4]= (mass / molar mass ) * volume L
So by substitution:
[Na2HPO4] = (8g/ 142g/mol) * 0.355 L = 0.02 M

and when Pka of the 2nd ionization of phosphoric acid = 7.21 
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PH = Pka + ㏒[A]/[AH]

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4 0
4 years ago
The reaction a(g)⇌2b(g) has an equilibrium constant of k = 0.030. what is the equilibrium constant for the reaction b(g)⇌12a(g)?
aalyn [17]
First, we have to correct the equation in the question to b(g)⇆ 1/2 A(g)
at the first equation A(g)⇆ 2 B(g) so,
 Kc = [B]^2 [ A] = 0.03 

by reverse the equation 2B⇆ A 
∴ Kc(original) = [A] / [B]^2

        = 1/0.03 = 33 M^-1

and the new equation B⇆ (1/2) A 
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7 0
4 years ago
Read 2 more answers
Methane is the main component of natural gas. Using the given reaction enthalpy, calculate the heat energy produced by the combu
SVETLANKA909090 [29]

Answer: The heat energy produced is 53831.25KJ

Explanation:

METHANE is the main component of natural gas. It can undergo combustion reaction in air with a bright blue flame to produce carbondioxide and water. The heat of reaction (enthalpy) is negative because heat is absorbed during the chemical reaction. To calculate the heat energy produced by the combustion of one kilogram (1 kg) of methane the following steps are taken:

Molecular mass of methane =16 gm/mol.

So moles of 1 kg methane =

Given mass of methane ÷ molecular weight of methane

But the given mass = 1kg = 1000g

Therefore,

moles of 1000g methane = 1000÷16

= 62.5 moles

Hence, energy evolved = (moles of methane) × (heat of combustion)

Therefore,

heat energy produced= 62.5 × (-861.3kj)

= -53831.25kj

​

7 0
3 years ago
15 POINTS RIGHT NOW!!!!
Anettt [7]

Answer:

It is valid because his experiment had clear variables.

Explanation:

He had clear variables and did everyhting the same for both plants, even though he had calcium in one and not in the other, which gives him a true and good result.

5 0
3 years ago
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How much energy is released by the decay of 3 grams of 230Th in the following reaction 230 Th - 226Ra + "He (230 Th = 229.9837 g
Serhud [2]

<u>Answer:</u> The energy released for the decay of 3 grams of 230-Thorium is 2.728\times 10^{-15}J

<u>Explanation:</u>

First we have to calculate the mass defect (\Delta m).

The equation for the alpha decay of thorium nucleus follows:

_{90}^{230}\textrm{Th}\rightarrow _{88}^{226}\textrm{Ra}+_2^{4}\textrm{He}

To calculate the mass defect, we use the equation:

Mass defect = Sum of mass of product - Sum of mass of reactant

\Delta m=(m_{Ra}+m_{He})-(m_{Th})

\Delta m=(225.9771+4.008)-(229.9837)=1.4\times 10^{-3}amu=2.324\times 10^{-30}kg

(Conversion factor: 1amu=1.66\times 10^{-27}kg )

To calculate the energy released, we use Einstein equation, which is:

E=\Delta mc^2

E=(2.324\times 10^{-30}kg)\times (3\times 10^8m/s)^2

E=2.0916\times 10^{-13}J

The energy released for 230 grams of decay of thorium is 2.0916\times 10^{-13}J

We need to calculate the energy released for the decay of 3 grams of thorium. By applying unitary method, we get:

As, 230 grams of Th release energy of = 2.0916\times 10^{-13}J

Then, 3 grams of Th will release energy of = \frac{2.0916\times 10^{-13}}{230}\times 3=2.728\times 10^{-15}J

Hence, the energy released for the decay of 3 grams of 230-Thorium is 2.728\times 10^{-15}J

5 0
3 years ago
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