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If 0.320 moles of zinc reacts with excess lead(IV) sulfate how many grams of zinc sulfate would be produced in the following reaction?
To determine the amount of zinc sulfate produced, we first need to look at the balanced chemical reaction between zinc and lead(IV) sulfate. The reaction is:
[ \text{Zn} + \text{Pb(SO}_4\text{)}_2 \rightarrow \text{ZnSO}_4 + \text{Pb} ]
From the reaction, 1 mole of zinc produces 1 mole of zinc sulfate (ZnSO₄). Therefore, 0.320 moles of zinc will produce 0.320 moles of zinc sulfate. To convert moles to grams, we multiply by the molar mass of zinc sulfate (approximately 161.44 g/mol), resulting in:
[ 0.320 , \text{moles} \times 161.44 , \text{g/mol} \approx 51.84 , \text{grams} ]
So, 51.84 grams of zinc sulfate would be produced.
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What is the following best describes a neutralization reaction?
An acid-base reaction that leaves no excess H+ or OH-
Why is excess magnesium is used for reacting it with sulphuric acid and powder?
Excess magnesium is used in reactions with sulfuric acid to ensure complete conversion of the acid into magnesium sulfate and hydrogen gas. The excess magnesium helps to drive the reaction to completion, allowing for efficient hydrogen gas production and preventing the formation of unreacted sulfuric acid. Additionally, using excess magnesium helps to control the reaction rate and mitigate the risk of localized overheating or reaction hazards.
How can you get crystals from the reaction of magnesium and sulphuric acid?
To obtain crystals from the reaction of magnesium and sulfuric acid, you can follow these steps: 1) Mix magnesium ribbon with diluted sulfuric acid in a beaker. 2) Allow the reaction to occur, which will produce hydrogen gas and magnesium sulfate. 3) Filter the mixture to remove any excess magnesium or impurities. 4) Evaporate the filtered solution to allow the magnesium sulfate to crystallize and form crystals.
Which ions is the limiting reagent and which is the reagent in excess when barium sulphate is made?
When barium sulfate is made, the limiting reagent is the one that is completely consumed in the reaction and determines the amount of product formed. In this case, if barium ions (Ba2+) and sulfate ions (SO42−) are the reactants, the limiting reagent would be the one that is present in lower molar quantity. The one in excess would be the one that is present in higher molar quantity. Without the quantities of each ion provided, it is difficult to determine which is the limiting reagent and which is in excess.
What would be need to do to produce solid magnesium sulphate?
To produce solid magnesium sulfate, you can start by reacting magnesium oxide (MgO) or magnesium hydroxide (Mg(OH)₂) with sulfuric acid (H₂SO₄). The reaction produces magnesium sulfate (MgSO₄) and water as a byproduct. To obtain solid magnesium sulfate, the resulting solution can be evaporated to remove excess water, allowing the magnesium sulfate to crystallize. The crystals can then be filtered and dried to yield solid magnesium sulfate.
Where is the excess copper from after adding zinc to copper sulfate?
The "excess" metallic copper produced by adding zinc metal to a copper sulfate solution comes from exchanging zinc atoms from the metal for copper atoms from the copper sulfate solution. During the reaction, the zinc atoms are ionized to cations and the copper cations from the solution are reduced to neutral atoms.
Why do we add an excess of copper carbonate to sulphuric acid when preparing the salt copper sulphate?
Adding an excess of copper carbonate ensures that all the sulfuric acid is fully neutralized and reacts with the copper carbonate to form copper sulfate. This guarantees that the maximum amount of copper sulfate is produced during the reaction.
What characteristics of a fission reaction makes a chain reaction possible?
The excess of neutrons produced.
When 400.0 g of ammonia reacted with excess sulfuric acid to produce ammonium sulfate 1463.0 g of product were obtained What is the percent yield of ammonium sulfate for this reaction?
94.25 - 94.28
When 300.0 g of ammonia reacted with excess sulfuric acid to produce ammonium sulfate 985.0 g of product were obtained What is the percent yield of ammonium sulfate for this reaction?
84.62 - 84.66 or 84.64%.
How many moles of copper will be produced if 5.8 mol of Zn react with excess copper II sulfate?
Also 5,8 mol
What is the following best describes a neutralization reaction?
An acid-base reaction that leaves no excess H+ or OH-
Why copper sulfate is added when acid is used up?
Copper sulfate is added after the acid is used up to test for any excess unreacted iron in a redox reaction. In the presence of excess iron, copper sulfate reacts with the iron to form a distinct blue color (copper precipitates as copper(I) oxide). This color change indicates the end point of the reaction, helping to determine when all the iron has been used up.
When excess of ammonia is added to aqueous zinc sulfate does the precipitate dissolve?
Yes. The ammonia will form ammonium hydroxide. The ammonium cation (NH4+) will react with SO4^2- to form the soluble salt ammonium sulfate, (NH4)2SO4. According to Le Chatelier's Principle, this will push the reaction to the right, thus forming more ammonium sulfate from the insoluble zinc sulfate.
How many moles of copper will be produced if 5.8 mol of zinc react with excess copper (ll) sulfate?
If zinc reacts with excess copper(ll) sulfate, a 1:1 molar ratio will be maintained. Therefore, 5.8 mol of zinc will produce 5.8 mol of copper.
When 500g of ammonia reacted with excess sulfuric acid to produce ammonium sulfate 1789g of product were obtained What is the percent yield of ammonium sulfate for this reaction?
The theoretical yield of ammonium sulfate can be calculated by determining the amount that would be produced if all the ammonia reacted. Given that 500g of ammonia was used, convert this amount to grams of ammonium sulfate. Then, divide the actual yield (1789g) by the theoretical yield and multiply by 100 to calculate the percent yield.
Add an excess of potassium iodide to copper sulphate solution how much is excess?
The amount of excess potassium iodide depends on the stoichiometry of the reaction between potassium iodide and copper sulfate. One equivalent of potassium iodide is needed to react with one equivalent of copper sulfate. Excess potassium iodide would be any amount added beyond this stoichiometric ratio.
