if 250 ml of 1.50 M HCl reacts with excess zinc how many moles of zinc...
Zinc reacts with hydrochloric acid according to the reaction equation Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g)Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g) How many milliliters of 2.00 M HCl(aq)2.00 M HCl(aq) are required to react with 8.55 g Zn(s)?
Zinc reacts with hydrochloric acid according to the reaction equation shown. Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g) How many milliliters of 5.00 M HCl(aq) are required to react with 2.65 g of an ore containing 26.0 % Zn(s) by mass? volume: mL
Zinc reacts with hydrochloric acid according to the reaction equation shown. Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g) How many milliliters of 6.50 M HCl(aq) are required to react with 4.45 g of an ore containing 39.0 % Zn(s) by mass?
Zinc reacts with hydrochloric acid according to the reaction equation shown. Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g) How many milliliters of 4.00 M HCl(aq) are required to react with 7.15 g of an ore containing 43.0 % Zn(s) by mass?
Zinc reacts with hydrochloric acid according to the reaction equation shown. Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g) How many milliliters of 5.50 M HCl(aq) are required to react with 2.95 g of an ore containing 43.0 % Zn(s) by mass?
Zinc reacts with hydrochloric acid to form hydrogen gas. A 0.258 g sample of Zinc reacts with 25.0 mL of 0.250 M HCI solution. What volume does the hydrogen has formed occupy at 24.0 °C and 0.990 atm? Zn(s) + 2HCl(aq) --ZnCl2(aq) + H2(g) 33.0 mL 97.3 mL 6.21 mL 77.0 mL
please help Zinc reacts with hydrochloric acid according to the reaction equation Zn(s) +2HCl(aq) ? ZnCl2(aq)-H2(g How many milliliters of 4.50 M HCl(aq) are required to react with 5.25 g of an ore containing 50.0% Zn(s) by mass? Number mL
Zinc reacts with hydrochloric acid according to the reaction equation shown. Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g) How many milliliters of 3.50M HCl(aq) are required to react with 2.65g of an ore containing 41.0% Zn(s) by mas
zinc metal reacts with excess hydrochloric acid to produce hydrogen gas according to the following equation: Zn(s) + 2HCl(aq)ZnCl2(aq) + H2(g) The product gas, H2, is collected over water at a temperature of 20 °C and a pressure of 745 mm Hg. If the wet H2 gas formed occupies a volume of 9.32 L, the number of moles of Zn reacted was __mol. The vapor pressure of water is 17.5 mm Hg at 20 °C.
Hydrogen gas can be prepared by reaction of zinc metal with aqueous HCl: Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g) How many liters of H2 would be formed at 680 mm Hg and 17 ∘C if 27.0 g of zinc was allowed to react? How many grams of zinc would you start with if you wanted to prepare 6.00 L of H2 at 400 mm Hg and 35.0 ∘C?