Question

Через порцию 25%го раствора гидроксида натрия массой 960грамм пропустили углекислый газ объемом 89,6 литров(н.у). Вычислите массовые доли солей в растворе после реакции.

Answer 1

2NaOH + CO2 = Na2CO3 + H2O 
NaOH + CO2 = NaHCO3 
в первом уравнении гидроксида натрия х моль, во втором y, составляем систему уравнений: 2 * 40х + 40 y = 240 (это 25% от 960) 
х + y = 4 (CO2 4 моль = 89,6/22,4) 
решаем, получаем х=2, y=2, значит соль в первом уравнении 106 * 2 = 212 г, и соль во втором уравнении 84*2=168; 
всего раствора получится 960 г + 44*4 CO2=1136, теперь находим масовые доли 212/1136*100=18,7 и 168/1136*100=14,8

Answer 2

Calculation of Mass Fractions of Salts in the Solution

To calculate the mass fractions of salts in the solution after the reaction, we need to determine the reactants and products involved in the reaction. From the given information, we know that a 25% solution of sodium hydroxide (NaOH) with a mass of 960 grams was reacted with carbon dioxide (CO2) gas with a volume of 89.6 liters at standard temperature and pressure (STP).

The balanced chemical equation for the reaction between sodium hydroxide and carbon dioxide is as follows:

2 NaOH + CO2 → Na2CO3 + H2O

According to the balanced equation, 2 moles of sodium hydroxide react with 1 mole of carbon dioxide to produce 1 mole of sodium carbonate (Na2CO3) and 1 mole of water (H2O).

To calculate the mass fractions of salts in the solution after the reaction, we need to determine the moles of reactants and products involved.

Moles of Sodium Hydroxide (NaOH)

To calculate the moles of sodium hydroxide, we can use the given mass and the molar mass of NaOH.

The molar mass of NaOH is the sum of the atomic masses of sodium (Na), oxygen (O), and hydrogen (H):

Na: 22.99 g/mol O: 16.00 g/mol H: 1.01 g/mol

Molar mass of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40.00 g/mol

The moles of sodium hydroxide can be calculated using the formula:

moles = mass / molar mass

moles of NaOH = 960 g / 40.00 g/mol = 24.00 mol

Moles of Carbon Dioxide (CO2)

To calculate the moles of carbon dioxide, we can use the ideal gas law equation:

PV = nRT

Where: P = pressure (STP = 1 atm) V = volume (89.6 L) n = moles of CO2 (to be calculated) R = ideal gas constant (0.0821 L·atm/(mol·K)) T = temperature (STP = 273.15 K)

Rearranging the equation to solve for moles:

n = PV / RT

n = (1 atm) * (89.6 L) / (0.0821 L·atm/(mol·K) * 273.15 K) = 3.47 mol

Moles of Sodium Carbonate (Na2CO3)

From the balanced chemical equation, we know that the mole ratio between sodium hydroxide and sodium carbonate is 2:1. Therefore, the moles of sodium carbonate formed will be half of the moles of sodium hydroxide reacted.

moles of Na2CO3 = 1/2 * moles of NaOH = 1/2 * 24.00 mol = 12.00 mol

Moles of Water (H2O)

From the balanced chemical equation, we know that the mole ratio between sodium hydroxide and water is 1:1. Therefore, the moles of water formed will be equal to the moles of sodium hydroxide reacted.

moles of H2O = moles of NaOH = 24.00 mol

Calculation of Mass Fractions

The mass fraction of a component in a solution is the ratio of the mass of that component to the total mass of the solution.

To calculate the mass fractions of sodium carbonate and water in the solution after the reaction, we need to determine their masses.

The molar mass of sodium carbonate (Na2CO3) is the sum of the atomic masses of sodium (Na), carbon (C), and oxygen (O):

Na: 22.99 g/mol C: 12.01 g/mol O: 16.00 g/mol

Molar mass of Na2CO3 = 2 * 22.99 g/mol + 12.01 g/mol + 3 * 16.00 g/mol = 105.99 g/mol

The mass of sodium carbonate can be calculated using the formula:

mass = moles * molar mass

mass of Na2CO3 = 12.00 mol * 105.99 g/mol = 1271.88 g

The mass of water is equal to the mass of sodium hydroxide reacted:

mass of H2O = 24.00 mol * 18.02 g/mol = 432.48 g

The total mass of the solution after the reaction is the sum of the masses of sodium carbonate and water:

total mass = mass of Na2CO3 + mass of H2O = 1271.88 g + 432.48 g = 1704.36 g

The mass fractions of sodium carbonate and water in the solution after the reaction can be calculated using the formula:

mass fraction = mass of component / total mass

mass fraction of Na2CO3 = 1271.88 g / 1704.36 g = 0.746 mass fraction of H2O = 432.48 g / 1704.36 g = 0.254

Therefore, the mass fractions of sodium carbonate and water in the solution after the reaction are approximately 0.746 and 0.254, respectively.