First construct a a balanced equation for the reaction :
2Na + 2H2O --> 2NaOH + H2
From the equation, we know that the ratio of Sodium(Na) to the ratio of Sodium Hydroxide(NaOH) is the same.
Ar of Na = 23g/mol
Number of moles = mass / Ar = 46g / 23g/mol = 2mol (moles of Na used)
Since ratio of Na to NaOH is the same, therefore there are also 2 moles of NaOH formed.
Yes, sodium hydroxide can react with copper to form copper(II) hydroxide. When sodium hydroxide is added to a solution containing copper ions, a blue precipitate of copper(II) hydroxide is formed.
The precipitate formed when barium chloride reacts with sodium hydroxide is barium hydroxide (Ba(OH)2).
Sodium chloride is formed when sodium hydroxide and hydrochloric acid react. This is a neutralization reaction where the sodium hydroxide (a base) and hydrochloric acid (an acid) combine to form a salt (sodium chloride) and water.
The balanced chemical equation for the reaction between sodium hydroxide and copper(III) chloride is: 2NaOH + 3CuCl3 → 3Cu(OH)3 + 6NaCl This equation shows that two moles of sodium hydroxide react with three moles of copper(III) chloride to produce three moles of copper(III) hydroxide and six moles of sodium chloride.
The reaction between sodium hydroxide and copper sulfate forms copper hydroxide and sodium sulfate. Copper hydroxide is initially formed as a blue precipitate, which can further react to form copper oxide upon heating.
In the acid-base reaction where sodium hydroxide and sulfuric acid react, the formula is: H2SO4 + 2NaOH --> Na2SO4 + 2H2O. The coefficients shown are necessary to uphold the law of conservation of mass. So, if you have 17 moles of sulfuric acid, you will need twice as many moles of sodium hydroxide, so the answer is 34 moles NaOH.
Yes, sodium hydroxide can react with copper to form copper(II) hydroxide. When sodium hydroxide is added to a solution containing copper ions, a blue precipitate of copper(II) hydroxide is formed.
The precipitate formed when barium chloride reacts with sodium hydroxide is barium hydroxide (Ba(OH)2).
Na +H2O -> NaOH +(1/2)H2 Every mole of Sodium requires one mole of water to make one mole of Sodium Hydroxide. So two moles of Sodium will produce two moles of Sodium Hydroxide. If there are three moles of water in the initial reaction then there will be one mole of water left over after reacting with two moles of Sodium. This reaction will produce half a mole of hydrogen gas.
Sodium chloride is formed when sodium hydroxide and hydrochloric acid react. This is a neutralization reaction where the sodium hydroxide (a base) and hydrochloric acid (an acid) combine to form a salt (sodium chloride) and water.
Sodium react with water and release hydrogen.Sodium hydroxide is also formed. Calcium hydroxide remain unchanged.
The balanced chemical equation for the reaction between sodium hydroxide and copper(III) chloride is: 2NaOH + 3CuCl3 → 3Cu(OH)3 + 6NaCl This equation shows that two moles of sodium hydroxide react with three moles of copper(III) chloride to produce three moles of copper(III) hydroxide and six moles of sodium chloride.
The reaction is: NaOH + HCl = NaCl + H2O HCl and NaCl cannot react, the anion is the same.
The reaction between sodium hydroxide and copper sulfate forms copper hydroxide and sodium sulfate. Copper hydroxide is initially formed as a blue precipitate, which can further react to form copper oxide upon heating.
The balanced equation for this reaction is: 2Na + 2H2O -> 2NaOH + H2. This means that two moles of sodium react with two moles of water to produce two moles of sodium hydroxide and one mole of hydrogen gas.
Sodium hydroxide and sodium sulfate don't actually react.
When 4 moles of sodium react with oxygen, 2 moles of sodium oxide are produced (according to the balanced equation). Therefore, when 10 moles of sodium react, 5 moles of sodium oxide will be produced.