Question

Решите пожалуйста!№1 Как можно получитьА) Cu(OH)дважды + Na2 SO4Б) Cu(OH)дважды + N2 O3В) CuO + SO2Г)СuO+H2 SO4№2 Какие пары веществ прореагируют ?А) Na2 Co3 + KClБ) Na2 Co3 + CaCl2B) Na2 Co3 + HClГ) Na2 Co3 + Mg№3 Вычислите объем оксида серы (IV), который образуется при растворении в соляной кислоте504 гсульфита натрия, содержащего 25% примесей?

Answer 1

Решите пожалуйста!
№1 Как можно получить ????? ЧТО именно? я допишу ур-ния, а ты сама смотри что))
А) Cu(OH)2 + Na2SO4 = CuSO4 + 2NaOH
Б) Cu(OH)2 + N2O3 = Сu(NO2)2 + H2O
В) CuO + SO2 = CuSO3
Г)СuO+H2SO4 = CusO4 + H2O

2) б,в

 

б)Na2CO3 + CaCl2 = CaCO3 + 2NaCl

в)Na2CO3 + 2HCl = 2NaCl + CO2 + H2O

Дано:                                                     

m (Na2SO3) = 504 г           

W(прим) = 25 %

Найти:

V (SO2)

Вычислим массу чистого Na2SO3:

w(чист) = 100% - w(прим) = 100% - 25% = 75%

504 г – 100%

х г   -  75%

х = 504 г ∙ 75% \100% = 378 г

реакция идет по двум уравнениям:

Na2SO3+ 2HCl = 2NaCl + Н2SO3

H2SO3 =  SO2 + H2O

Суммарное: Na2SO3+ 2HCl = 2NaCl + SO2 + H2O

Вычислим объем выделившегося SO2:

 378 г                                   х л

Na2SO3+ 2HCl = 2NaCl + SO2 + H2O

 126 г                                  22,4  г

378  г   ---   х г

126 г  ---   22,4 л

х = 378 г ∙ 22,4 л\126 г = 67,2 л

ответ:  V (SO2) = 67,2 л

 

 

 

 

Answer 2

Answer 1:

To obtain Cu(OH)2 from Na2SO4, you can perform a double displacement reaction between copper sulfate (CuSO4) and sodium hydroxide (NaOH). The balanced equation for this reaction is:

CuSO4 + 2NaOH -> Cu(OH)2 + Na2SO4

In this reaction, copper sulfate reacts with sodium hydroxide to form copper hydroxide and sodium sulfate. The state symbols for the reactants and products are not provided, so it is assumed that they are in their standard states.

Answer 2:

Let's determine which pairs of substances will react:

A) Na2CO3 + KCl: This is a double displacement reaction between sodium carbonate and potassium chloride. The balanced equation is:

Na2CO3 + 2KCl -> 2NaCl + K2CO3

B) Na2CO3 + CaCl2: This is also a double displacement reaction between sodium carbonate and calcium chloride. The balanced equation is:

Na2CO3 + CaCl2 -> 2NaCl + CaCO3

C) Na2CO3 + HCl: This is a double displacement reaction between sodium carbonate and hydrochloric acid. The balanced equation is:

Na2CO3 + 2HCl -> 2NaCl + H2CO3

D) Na2CO3 + Mg: Sodium carbonate and magnesium do not react with each other.

Therefore, the pairs of substances that will react are A) Na2CO3 + KCl, B) Na2CO3 + CaCl2, and C) Na2CO3 + HCl.

Answer 3:

To calculate the volume of sulfur dioxide (SO2) formed when dissolving 504 g of sodium sulfite (Na2SO3) containing 25% impurities in hydrochloric acid (H2SO4), we need to consider the balanced equation for the reaction between sodium sulfite and hydrochloric acid:

Na2SO3 + H2SO4 -> Na2SO4 + H2O + SO2

From the balanced equation, we can see that 1 mole of sodium sulfite reacts with 1 mole of hydrochloric acid to produce 1 mole of sulfur dioxide. To calculate the number of moles of sodium sulfite, we need to divide the mass of sodium sulfite by its molar mass:

Molar mass of Na2SO3 = 2(22.99 g/mol) + 32.06 g/mol + 3(16.00 g/mol) = 126.04 g/mol

Number of moles of Na2SO3 = 504 g / 126.04 g/mol = 4 mol

Since the sodium sulfite contains 25% impurities, we need to consider only 75% of the moles of sodium sulfite:

Number of moles of Na2SO3 (after considering impurities) = 4 mol * 0.75 = 3 mol

Therefore, 3 moles of sodium sulfite will produce 3 moles of sulfur dioxide. To calculate the volume of sulfur dioxide, we need to use the ideal gas law:

PV = nRT

Assuming the reaction takes place at standard temperature and pressure (STP), we can use the values: - Pressure (P) = 1 atm - Temperature (T) = 273 K - Gas constant (R) = 0.0821 L·atm/(mol·K)

Rearranging the equation, we can solve for volume (V):

V = nRT / P = 3 mol * 0.0821 L·atm/(mol·K) * 273 K / 1 atm = 6.68 L

Therefore, the volume of sulfur dioxide formed is approximately 6.68 liters.