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kykrilka [37]
3 years ago
7

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

ow your solution and explain how you derive the final answer
Chemistry
2 answers:
a_sh-v [17]3 years ago
5 0
We are given with the equation 2Na + O2 → 2Na2O where Na and O2 react to form sodium oxide. We are also given with two amounts 55.3 g Na with 64.3 g O2 .In this case, we will find the limiting reactant. we divide each with the molar mass and the stoich coeff. Na is 1.202, O2 is 2.01. Hence the limiting is Na. we base the calculations here.The amount is equal to 46.88 grams
GenaCL600 [577]3 years ago
3 0
 <span>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</span>
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A sucrose solution is prepared to a final concentration of 0.250 M . Convert this value into terms of g/L, molality, and mass %.
MArishka [77]

Answer:

A. 85.6 g

= 0.0856 kg.

B. 0.00027 mol/g

= 0.27 mol/kg.

C. 8.39 %

Explanation:

Given:

Molar concentration = 0.25 M

Molar weight of sucrose = 342.296 g/mol

Density of solution = 1.02 g/mL

Mass of water = 934.4 g.

Density in g/l = 1.020 g/ml * 1000ml/1 l

= 1020 g/l

Mass of solution in 1 l of solution = 1020 g

Mass of solution = mass of solvent + mass of solute

Mass of sucrose = 1020 - 934.4

= 85.6 g of sucrose in 1 l of solution.

A.

Density of sucrose = mass/volume

= molar mass/molar concentration

= 342.296 * 0.25

= 85.6 g/l

Number of moles = mass/molar mass

= 85.6/342.296

= 0.25 mol

B.

Molality = number of moles of solute/mass of solvent

= 0.25/934.4

= 0.00027 mol/g

C.

% mass of sucrose = mass of sucrose/total mass of solution * 100

= 85.6/1020 * 100

= 8.39 %

6 0
3 years ago
How many moles of each element are in one mole of Be(OH)2?
REY [17]
B. 1 mole of beryllium, 2 moles of oxygen, 2 moles of hydrogen
8 0
3 years ago
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What property of water molecules allows water to dissolve many substances
hodyreva [135]

Answer:

Its high polarity

Explanation:

Solvent is the that part of solution which is present in large proportion and have ability to dissolve the solute. In simplest form it is something in which other substance get dissolve. The most widely used solvent is water, other examples are toluene, acetone, ethanol, chloroform etc.

Water is called universal solvent because of high polarity all polar substance are dissolve in it. Hydrogen is less electronegative while oxygen is more electronegative and because of difference in electronegativity hydrogen carry the partial positive charge while oxygen carry partial negative charge.

Water create electrostatic interaction with other polar molecules. The negative end of water attract the positive end of polar molecules and positive end of water attract negative end of polar substance and in this way polar substance get dissolve in it.

Example:

when we stir the sodium chloride into water the cation Na⁺ ions are surrounded by the negative end of water i.e oxygen and anion Cl⁻ is surrounded by the positive end of water i.e hydrogen and in this way all salt is get dissolved.

5 0
3 years ago
Consider the reaction between iron (III) oxide, Fe2O3 and carbon monoxide, CO.
Lana71 [14]
<span>Fe2O3 + 3CO --> 2Fe + 3CO2
</span><span>
m(Fe2O3)=213 g
m(CO)=140 g
</span>_______________

<span>n(Fe2O3)=?
m(Fe)=?
n(Fe2O3)=?
n(CO)=?
n(CO2)=?
</span>
<span>n(Fe2O3)=m(Fe2O3) / M(Fe2O3)
n(Fe2O3)= 213 g / 159,7 gmol-1 = 1,33 mol
</span>
<span>n(CO)= m(CO) / M(CO) n(CO)= 140 g / 28,01 gmol-1 = 4,99 mol</span>
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3 years ago
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3 years ago
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