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3 years ago

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:

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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

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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.

1) What is the partial pressure of SO₂ gas in the larger container?

∵ P = nRT/V.

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.

2) What is the partial pressure of N₂ gas in the larger container?

∵ P = nRT/V.

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.

3) What is the total pressure in the vessel?

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

∵ The total pressure in the vessel = the partial pressure of SO₂ + the partial pressure of N₂.

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

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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 Chlorine Cl = 35.45 g (See in group 17)

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 Avogadro's number,
1 mole = 1 * $6.022 * 10^{23}$ 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

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