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Alinara [238K]
3 years ago
11

If 38g of sodium hydroxide reacts with 28g of carbon, how many grams of sodium carbonate are produced?

Chemistry
1 answer:
scZoUnD [109]3 years ago
5 0

Answer:

Mass of sodium carbonate = 33.92 g

Explanation:

Given data:

Mass of sodium hydroxide = 38 g

Mass of carbon = 28 g

Mass of sodium carbonate produced = ?

Solution:

Chemical equation:

6NaOH + 2C → 2Na + 3H₂ + 2Na₂CO₃

Now will calculate the number of moles of reactants:

Number of moles of sodium hydroxide:

Number of moles = mass / molar mass

Number of moles = 38 g/ 40 g/mol

Number of moles = 0.95 mol

Number of moles of carbon:

Number of moles = mass / molar mass

Number of moles = 28 g/ 16 g/mol

Number of moles = 1.75 mol

Now we will compare the moles of both reactant with sodium carbonate.

                         C           :         Na₂CO₃

                         2           :           2

                       1.75         :         1.75

                  NaOH          :       Na₂CO₃      

                    6               :           2

                     0.95       :           2/6×0.95 = 0.32  

Number of moles of sodium carbonate produced by sodium hydroxide are less so it will limiting reactant.

Mass of sodium carbonate:

Mass = number of moles × molar mass

Mass = 0.32 mol × 106 g/mol

Mass = 33.92 g

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Answer:

-0.050 kJ/mol.K

Explanation:

  • A certain reaction is thermodynamically favored at temperatures below 400. K, that is, ΔG° < 0 below 400. K
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All in all, ΔG° = 0 at 400. K.

We can find ΔS° using the following expression.

ΔG° = ΔH° - T.ΔS°

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What is the symbol for the carbon isotope with seven neutrons?
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When one form of energy is transformed into another no energy is destroyed in the process?
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3 years ago
a sample of 3.00 g of so2 (g)originally in a 5.00 L vesselat 21 degee Celsius is transferred to a 10.0 L vessel at 26 degree Cel
eimsori [14]

Answer:

1) The partial pressure of SO₂ gas in the larger container = 0.115 atm.

2) The partial pressure of N₂ gas in the larger container = 0.206 atm.

3) The total pressure in the vessel = 0.321 atm.

Explanation:

  • To calculate the partial pressure of each gas, we can use the general law of ideal gas: PV = nRT.

where, P is the partial pressure of the gas in atm,

V is the volume of the vessel in L,

n is the no. of moles of the gas,

R is the general gas constant (R = 0.082 L.atm/mol.K),

T is the temperature of the gas in K.

<u><em>1) What is the partial pressure of SO₂ gas in the larger container?</em></u>

<em>∵ P = nRT/V.</em>

n = mass/molar mass = (3.0 g)/(64.066 g/mol) = 0.047 mol.

R = 0.082 L.atm/mol.K.

T = 26 °C + 273.15 = 299.15 K.

V = 10.0 L. (The volume of the new container)

∴ P = nRT/V = (0.047 mol)(0.082 L.atm/mol.K)(299.15 K)/(10.0 L) = 0.115 atm.

<u><em>2) What is the partial pressure of N₂ gas in the larger container?</em></u>

<em>∵ P = nRT/V.</em>

n = mass/molar mass = (2.35 g)/(28.0 g/mol) = 0.084 mol.

R = 0.082 L.atm/mol.K.

T = 26 °C + 273.15 = 299.15 K.

V = 10.0 L. (The volume of the new container)

∴ P = nRT/V = (0.084 mol)(0.082 L.atm/mol.K)(299.15 K)/(10.0 L) = 0.206 atm.

<u><em>3) What is the total pressure in the vessel?</em></u>

  • According to Dalton's law the total pressure exerted is equal to the sum of the partial pressures of the individual gases.

<em>∵ The total pressure in the vessel = the partial pressure of SO₂ + the partial pressure of N₂.</em>

∴ The total pressure in the vessel = 0.115 + 0.206 = 0.321 atm.

5 0
3 years ago
How many molecules of CaCl2 are equivalent to 75.9 g CaCl2
sergij07 [2.7K]
First, you need to find:
One mole of CaCl_{2} is equivalent to how many grams?

Well, for this you have to look up the periodic table. According to the periodic table:
The atomic mass of Calcium Ca = 40.078 g (See in group 2)
The atomic mass of <span>Chlorine Cl = 35.45 g (See in group 17)
</span>
As there are two atoms of Chlorine present in CaCl_{2}, therefore, the atomic mass of CaCl_{2} would be:

Atomic mass of CaCl_{2}  = Atomic mass of Ca + 2 * Atomic mass of Cl

Atomic mass of CaCl_{2} = 40.078 + 2 * 35.45 = 110.978 g

Now,

110.978 g of CaCl_{2} = 1 mole.
75.9 g of CaCl_{2} = \frac{75.9}{110.978} moles = 0.6839 moles.

Hence,
The total number of moles in 75.9g of CaCl_{2} = 0.6839 moles

According to <span>Avogadro's number,
1 mole = 1 * </span>6.022 * 10^{23} molecules
0.6839 moles = 0.6839 * 6.022 * 10^{23} molecules = 4.118*10^{23} molecules

Ans: Number of molecules in 75.9g of  CaCl_{2} =  4.118*10^{23} molecules

-i


3 0
3 years ago
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