Question

1Даны вещества: Оксид цинка Фосфор Вода Соляная кислота.Получите Цинк,оксид фосфора,Водород,хлорид цинка.2. 27г оксида олова(2) востоновили водородом.Определите объем водорода необходимый для реакции и массу полученного олова?

Answer 1

1. К - элемент I группы главной подгруппы, у него ВСЕГДА на внешнем уровне 1 электрон, соответственно и степень окисления +1 (ему легче этот электрон отдать, чем 7 электронов принять). К ВСЕГДА обладает свойствами щелочных металлов (это типичный металл), значит он может образовывать ТОЛЬКО основные оксиды, формула высшего оксида К2О (общая формула высших оксидов находится в самом низу таблицы Менделеева под каждой группой, там вместо R в формуле подставляете элемент нужный, а все индексы сохраняете) 
Соответственно, свойства основных оксидов: 
а) реакция с водой 
К2О + Н2О = 2КОН 
б) с кислотами 
К2О + Н2SO4 = K2SO4 + H2O 
в) с амфотерными основаниями 
3К2О+ 2Al(OH)3 = 2K3AlO3 + 3H2O 
г) с кислотными оксидами 
К2O + CO2 = K2CO3 
2. гидроксид калия - щелочь (тривиальное название - едкое кали) - КОН

Answer 2

Chemical Reactions and Calculations

To obtain zinc, zinc oxide, hydrogen, and zinc chloride from the given substances (zinc oxide, phosphorus, water, and hydrochloric acid), we need to perform specific chemical reactions. Let's break down the process step by step.

1. Obtaining Zinc (Zn): - Zinc can be obtained by reacting zinc oxide (ZnO) with carbon (C) in a process called reduction. - The chemical equation for this reaction is: ZnO + C → Zn + CO. - The carbon reduces the zinc oxide to produce zinc and carbon monoxide gas. - The reaction can be carried out by heating a mixture of zinc oxide and carbon in a furnace. - The zinc vaporizes and is collected as a solid upon cooling.

2. Obtaining Zinc Oxide (ZnO): - Zinc oxide can be obtained by heating zinc metal in the presence of oxygen. - The chemical equation for this reaction is: 2Zn + O2 → 2ZnO. - When zinc metal reacts with oxygen gas, zinc oxide is formed.

3. Obtaining Hydrogen (H2): - Hydrogen gas can be obtained by reacting water (H2O) with a reactive metal, such as zinc (Zn). - The chemical equation for this reaction is: Zn + 2H2O → Zn(OH)2 + H2. - Zinc displaces hydrogen from water, producing zinc hydroxide and hydrogen gas. - The hydrogen gas can be collected by upward displacement of water or using other methods.

4. Obtaining Zinc Chloride (ZnCl2): - Zinc chloride can be obtained by reacting zinc metal (Zn) with hydrochloric acid (HCl). - The chemical equation for this reaction is: Zn + 2HCl → ZnCl2 + H2. - Zinc reacts with hydrochloric acid to produce zinc chloride and hydrogen gas.

Now, let's move on to the second part of the question.

5. Calculating the Volume of Hydrogen and the Mass of Tin: - To determine the volume of hydrogen gas required for the reaction, we need to use the concept of stoichiometry. - The balanced chemical equation for the reaction between tin oxide (SnO2) and hydrogen gas (H2) is: SnO2 + 2H2 → Sn + 2H2O. - According to the equation, 1 mole of tin oxide reacts with 2 moles of hydrogen gas to produce 1 mole of tin and 2 moles of water. - The molar mass of tin oxide (SnO2) is 150.71 g/mol. - The given mass of tin oxide is 27 g. - To calculate the number of moles of tin oxide, we divide the mass by the molar mass: 27 g / 150.71 g/mol = 0.1794 mol. - Since the stoichiometric ratio between tin oxide and hydrogen gas is 1:2, we need twice the number of moles of hydrogen gas: 2 * 0.1794 mol = 0.3588 mol. - The molar volume of a gas at standard temperature and pressure (STP) is 22.4 L/mol. - Therefore, the volume of hydrogen gas required for the reaction is: 0.3588 mol * 22.4 L/mol = 8.03 L. - Finally, to determine the mass of tin produced, we need to use the molar mass of tin (Sn), which is 118.71 g/mol. - The mass of tin can be calculated by multiplying the number of moles of tin by its molar mass: 0.1794 mol * 118.71 g/mol = 21.32 g.

Please note that the calculations provided are based on the given information and assumptions about the reaction conditions. It is always important to double-check the calculations and consider any additional factors that may affect the reaction.