Non!!!! They are two different substances/acids.
The formulas for sulphuric acid is H2SO4 and for hydrochloric acid (HCl).
With the exception of the hydrogens all the other elements present in the two difference acids are different.
Balanced equation. Mg + H2SO4 -> MgSO4 + H2 ------------------------------------- ( sufuric acid in excess---magnesium limits, drives reaction ) 24 grams magnesium (1 mole Mg/24.31 g)(1 mole H2/1 mole Mg)(2.016 g/1 mole H2) = 1.9 grams of hydrogen gas produced ( can call it 2.0 grams ) --------------------------------------------------
To determine the number of grams of Cu(NO3)2 produced, you need to consider the molar ratio between Cu(NO3)2 and Cu. Firstly, convert the 4.2 grams of Cu to moles using the molar mass of Cu. Then, use the balanced chemical equation to find the moles of Cu(NO3)2 produced. Finally, convert the moles of Cu(NO3)2 to grams using its molar mass.
To find the molarity, start by converting grams of sulfuric acid to moles. the molar mass of sulfuric acid (H2SO4) is 98.08 g/mol. Divide the moles by the volume in liters (500.0 mL = 0.5 L) to get the molarity. Molarity = moles/volume (L).
The "hydro" part indicates there's hydrogen in it, but not all hydrogen-containing acids start with hydro...sulfuric acid contains more hydrogen than either hydrochloric or hydrofluoric. "-ic" is pretty common for acids...hydrochloric, hydrofluoric, sulfuric, nitric, acetic...
Lithium chloride is not transformed in calcium chloride.
Balanced equation. Mg + H2SO4 -> MgSO4 + H2 ------------------------------------- ( sufuric acid in excess---magnesium limits, drives reaction ) 24 grams magnesium (1 mole Mg/24.31 g)(1 mole H2/1 mole Mg)(2.016 g/1 mole H2) = 1.9 grams of hydrogen gas produced ( can call it 2.0 grams ) --------------------------------------------------
The chemical equation is:2 NaOH + H2SO4 = Na2SO4 + 2 H2OMolar mass of sodium hydroxide is 39,9971 g; molar mass of sulfuric acid is 98,079 g.2 . 39,9971 g NaOH----------------------98,079 g H2SO4200 g NaOH------------------------xx = (200 x 98,079)/2 . 39,9971 = 245 g H2SO4So sulfuric acid is the limiting reagent.
To determine the number of grams of Cu(NO3)2 produced, you need to consider the molar ratio between Cu(NO3)2 and Cu. Firstly, convert the 4.2 grams of Cu to moles using the molar mass of Cu. Then, use the balanced chemical equation to find the moles of Cu(NO3)2 produced. Finally, convert the moles of Cu(NO3)2 to grams using its molar mass.
To find the molarity, start by converting grams of sulfuric acid to moles. the molar mass of sulfuric acid (H2SO4) is 98.08 g/mol. Divide the moles by the volume in liters (500.0 mL = 0.5 L) to get the molarity. Molarity = moles/volume (L).
6,36 g of silver chloride are obtained.
Lithium chloride is not transformed in calcium chloride.
The "hydro" part indicates there's hydrogen in it, but not all hydrogen-containing acids start with hydro...sulfuric acid contains more hydrogen than either hydrochloric or hydrofluoric. "-ic" is pretty common for acids...hydrochloric, hydrofluoric, sulfuric, nitric, acetic...
To calculate the grams of iron II sulfide needed, we start by finding the moles of hydrogen sulfide produced. This is done by dividing the given mass of hydrogen sulfide by its molar mass. Then, we use the balanced chemical equation to determine that for every 4 moles of hydrogen sulfide, 1 mole of iron II sulfide is needed. From this, we find the grams of iron II sulfide required by multiplying the moles of iron II sulfide by its molar mass.
735 g of Ca3(PO4)2 are obtained.
Following the Law of Conservation of Mass (see link below), there will be 20 grams of products in a reaction of 20 grams of reactions.
Balanced equation. 2NaOH + H2SO4 -> Na2SO4 + 2H2O 12.5 grams NaOH (1 mole NaOH/39.998 grams)(1 mole Na2SO4/2 mole NaOH)(142.05 grams/1 mole Na2SO4) = 22.2 grams sodium sulfate produced ===========================
If you start with 20.8 grams of barium sulfate (BaSO₄), you cannot produce more barium sulfate from it; you can only measure how much you have. Therefore, you can produce a maximum of 20.8 grams of barium sulfate if you are referring to using the same amount of BaSO₄ in a reaction or process. In summary, you have 20.8 grams of barium sulfate available, not more.