Should be, assuming ideal reaction.......
Pb + H2SO4 --> PbSO4 + H2
Products are lead sulfate and hydrogen gas.
KBrO3 and KBr do not react with each other because they have same cation K+, but separately they react with H2SO4. 2KBrO3 + H2SO4 -------> K2SO4 + 2HBrO3 2KBr + H2SO4 ----------> K2SO4 + 2HBr
When litharge (lead(II) oxide) is heated with concentrated sulfuric acid (H2SO4), lead sulfate (PbSO4) is formed along with water. The reaction is as follows: PbO(s) + H2SO4(l) → PbSO4(s) + H2O(l)
2 moles of NaOH will react with 1 mole of H2SO4 based on the balanced chemical equation: 2NaOH + H2SO4 -> Na2SO4 + 2H2O.
In theory, lead should react with dilute sulfuric acid, to give lead(II) sulfate and hydrogen. In practice the reaction is very slow, because lead sulfate is insoluble in water and quickly clogs up the lead surface, and because lead is relatively unreactive.
When lead(II) nitrate (Pb(NO3)2) and sulfuric acid (H2SO4) react, lead(II) sulfate (PbSO4) and nitric acid (HNO3) are produced. This reaction is a double displacement reaction where the ions in the reactants switch partners to form the products. Lead(II) sulfate is an insoluble salt that precipitates out of solution.
When lead (II) nitrate is mixed with sulfuric acid (H2SO4), a white precipitate of lead sulfate (PbSO4) is formed along with nitric acid as a byproduct. This reaction is a double displacement reaction where the lead ions from the nitrate salt react with sulfate ions from sulfuric acid to form the insoluble lead sulfate.
KBrO3 and KBr do not react with each other because they have same cation K+, but separately they react with H2SO4. 2KBrO3 + H2SO4 -------> K2SO4 + 2HBrO3 2KBr + H2SO4 ----------> K2SO4 + 2HBr
When litharge (lead(II) oxide) is heated with concentrated sulfuric acid (H2SO4), lead sulfate (PbSO4) is formed along with water. The reaction is as follows: PbO(s) + H2SO4(l) → PbSO4(s) + H2O(l)
NaOH can react with H2SO4 to form Na2SO4 and water. Ca(OH)2 can react with H2SO4 to form CaSO4 and water. HF can react with H2SO4 to form H2SO4 + HF -> H2O + SO3 + HF + HF, as well as AlCl3O2H2O can react with H2SO4 to form Al2(SO4)3 +H2O. Mg(OH)2 can react with H2SO4 to form MgSO4 and water.
2 moles of NaOH will react with 1 mole of H2SO4 based on the balanced chemical equation: 2NaOH + H2SO4 -> Na2SO4 + 2H2O.
In theory, lead should react with dilute sulfuric acid, to give lead(II) sulfate and hydrogen. In practice the reaction is very slow, because lead sulfate is insoluble in water and quickly clogs up the lead surface, and because lead is relatively unreactive.
When lead(II) nitrate (Pb(NO3)2) and sulfuric acid (H2SO4) react, lead(II) sulfate (PbSO4) and nitric acid (HNO3) are produced. This reaction is a double displacement reaction where the ions in the reactants switch partners to form the products. Lead(II) sulfate is an insoluble salt that precipitates out of solution.
Let's see. MgCO3 + H2SO4 --> MgSO4 + CO2 + H2O A salt, magnesium sulfate, carbon dioxide and water are produced in this reaction.
Lead can react with chloride salts to form insoluble lead chloride (PbCl2), such as in the reaction with hydrochloric acid (HCl) to form lead(II) chloride (PbCl2). Lead can also react with sulfate salts to form insoluble lead sulfate (PbSO4), as in the reaction with sulfuric acid (H2SO4) to form lead(II) sulfate (PbSO4).
When aqueous H2SO4 and aqueous NaOH react by double replacement, they form water (H2O) and sodium sulfate (Na2SO4) as products. The reaction is H2SO4 + 2NaOH → 2H2O + Na2SO4.
When lead nitrate is reacted with sulfuric acid, lead sulfate and nitric acid are formed. The balanced chemical equation for this reaction is: Pb(NO3)2 + H2SO4 -> PbSO4 + 2HNO3. This reaction is a double displacement type.
The newly formed lead compound will be lead chlorate. Lead sulfate will react with tin chlorate to form lead chlorate as a product.