4Al(s) +302(g) ----> 2Al2O3(s)
When 4 moles of aluminum react with an excess of chlorine gas, 4 moles of aluminum chloride are produced. This is because the balanced chemical equation for the reaction is: 2Al + 3Cl2 -> 2AlCl3 This means that 2 moles of aluminum react with 3 moles of chlorine gas to produce 2 moles of aluminum chloride, so 4 moles of aluminum will produce 4 moles of aluminum chloride.
Al(OH)3 + 3NaNO3 >> Al(NO3)3 + 3NaOH This is a double displacement reaction between metals ( Aluminum is a transition metal, but has only a 3+ oxidation state. ) and polyatomic ions of negative charge. Both are 1- in charge. Sodium is 1+ in charge. This is why the atoms are arranged as they are with the polyatomic ions enclosed in parentheses where needed. Remember, botgh sides of the equation musr contain the same number of atoms. Count carefully and you will see this is the case.
The balanced chemical equation for the reaction between sulfurous acid and aluminum hydroxide to form aluminum sulfite and water is: 3H2SO3 + 2Al(OH)3 → Al2(SO3)3 + 6H2O This equation shows that 3 moles of sulfurous acid and 2 moles of aluminum hydroxide react to produce 1 mole of aluminum sulfite and 6 moles of water.
The balanced chemical equation for the reaction between aluminum and chlorine is 2Al + 3Cl2 -> 2AlCl3. This means that for every 2 moles of aluminum that react, 2 moles of aluminum chloride are produced. Therefore, if 0.440 mol of aluminum is used, it will produce 0.440 mol of aluminum chloride.
3 O2(g) + 2 S(s) --> 2 SO3(g) (This turns into liquid below 45°C)
The balanced equation for aluminum reacting with iron(II) oxide to produce aluminum oxide and iron is: 2Al + FeO -> Al2O3 + 2Fe.
The balanced chemical equation for potassium phosphate (K3PO4) reacting with aluminum nitrate (Al(NO3)3) to produce potassium nitrate (KNO3) and aluminum phosphate (AlPO4) is: 2K3PO4 + 3Al(NO3)3 → 3KNO3 + AlPO4
Based on the balanced chemical equation provided, 3 molecules of water are formed for every 2 molecules of aluminum oxide reacting with 6 molecules of hydrochloric acid to produce 2 molecules of aluminum chloride and the 3 molecules of water.
The balanced chemical equation for the reaction between aluminum and oxygen to form aluminum oxide is: 4Al + 3O2 → 2Al2O3. From the balanced equation, we can see that 4 moles of aluminum react with 3 moles of oxygen to produce 2 moles of aluminum oxide. Therefore, if 3.40 mol of aluminum and 2.85 mol of oxygen are reacted, the limiting reactant is oxygen. Thus, 3.40 mol of aluminum would theoretically produce 1.90 mol of aluminum oxide.
To balance the equation for the reaction of aluminum (Al) with oxygen (O2) to form aluminum oxide (Al2O3), we start with the unbalanced equation: 4Al + 3O2 → 2Al2O3. The coefficient on Al is 4 in this balanced equation, indicating that four aluminum atoms are needed to react with three molecules of oxygen to produce two formula units of aluminum oxide.
When aluminum carbide reacts with water, the products of the reaction are aluminum hydroxide and methane gas. The balanced equation for this reaction is Al4C3 + 12H2O -->4Al(OH)3 + 3CH4(g)
3Ni(NO2)2 (aq) + 2Na3PO4 (aq) = Ni3(PO4)2 (s) + 6NaNO3 (aq)
The chemical equation is:2 AlBr3 + 3 Cl2 = 2 AlCl3 + 3 Br2
If you have 30 molecules of sodium reacting with aluminum chloride (AlCl3), they will produce 10 molecules of aluminum. This is because the balanced equation shows that 3 moles of sodium react with 1 mole of aluminum, producing 1 mole of aluminum.
To find the amount of copper (II) sulfate needed to react with 0.48 mol of aluminum (III) sulfate, start by writing a balanced chemical equation for the reaction between the two salts. From the balanced equation, determine the molar ratio between copper (II) sulfate and aluminum (III) sulfate. Then, use this ratio to calculate the amount of copper (II) sulfate needed to produce 0.48 mol of aluminum (III) sulfate.
In this reaction, aluminum metal reacts with sulfuric acid to produce aluminum sulfate solution and hydrogen gas. The balanced chemical equation for this reaction is: 2Al + 3H2SO4 -> Al2(SO4)3 + 3H2.
The stoichiometry for making aluminum oxide (Al2O3) involves the reaction between aluminum metal and oxygen gas. The balanced chemical equation is 4Al + 3O2 -> 2Al2O3, which means that 4 moles of aluminum react with 3 moles of oxygen gas to produce 2 moles of aluminum oxide.