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How many moles of HCl and needed to react 2.4 moles of Al?


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Answered 2016-02-25 08:24:47

Al+HCl===> AlCl3+H2 Is the reaction. You need &.2 moles of HCl.

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Answered 2015-07-23 20:33:43

The answer is 7, moles HCl.

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This reaction emmits H2 gas. We need 7.2 moles of Hcl.


2 AL + 6HCl -> 2AlCl3 + 3H2 Three times as much HCl is needed by unit to react with the Al. 2.4 * 3 = 7.2 7.2 moles of HCl



The reaction between HCl & Al will be follow as : 6HCl + 2Al = 2AlCl3 + 3H2 Now 2 mole of Al reacts with = 6 moles of HCl Then 0.06591 moles of Al will react with = 3X0.06591 = 0.19773 mole of HCl


I assume you mean this reaction. Zn + 2HCl --> ZnCl2 + H2 2.3 moles zinc (2 moles HCl/1 mole Zn) = 4.6 moles hydrochloric acid needed ========================


molar mass of NaHCO3 =84 moles of nahco3 =0.35/84 = 4.2*10^-3 moles with balanced equation{ HCl + NaHCO3 ---> NaCl + H2O + CO2}, one mole of HCl react with one mole of NaHCO3 but 4.2*10^-3 moles of NaHCO3 react with =4.2*10^-3*1 mol hcl=4* 10^-3 moles


How many moles of CO2 form when 15.5 ml of 3.00m HCl solution react?


The chemical reaction isȘTiCl4 + 2 H2O = TiO2 + 2 HCl13 moles water are needed.


Taking rust to be Fe2O3, you would have the following reaction:Fe2O3 + 6HCl ==> 2FeCl3 + 3H2O100 g Fe2O3 x 1 mole Fe2O3/159.7 g = 0.626 moles Fe2O3moles HCl needed = 0.626 moles Fe2O3 x 6 moles HCl/mole Fe2O3 = 3.76 moles HCl neededMass HCl needed = 3.76 moles HCl x 36.5 g/mole = 137 g HCl needed


Use grams to moles to moles to grams: 0.2 g of ammonia gas (NH3) is equivalent to 0.012 moles of NH3 (divide by 17g/mole) One mole of NH3 reacts with one mole of HCl: NH3 + HCl <=> NH4Cl So we need 0.012 moles of HCl to react with 0.012 moles of NH3 0.012 moles HCl * 36.5 g/mole HCl => 0.43 g HCl


Balanced equation always first!! Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O Find moles HCl by---Molarity = moles of solute/Liters of solution (18.1 ml = 0.0181 L) 0.23 M HCl = moles HCl/0.0181 liters = 0.004163 moles HCl Drive back against sodium carbonate stoichometrically 0.004163 moles HCl (1 mole Na2CO3/2 mole HCl)(105.99 grams/1 mole Na2CO3) = 0.22 grams Na2CO3 needed --------------------------------------------


First.Molarity = moles of solute/Liters of solution0.22 M HCl = X moles/1.0 L= 0.22 moles HCl--------------------------Second.0.22 moles HCl (36.458 grams/1 mole HCl)= 8.0 grams hydrochloric acid needed===========================


Balanced equation first. CaCO3 + 2HCl -> CaCl2 + CO2 + H2O 2.5 moles CaCO3 (2 mole HCl/1 mole CaCO3) = 5.0 moles HCl required ==================


Na2CO3+2HCl-->2NaCl + CO2 ratio is 1:2.So 0.00265 x 2 mol of HCl needed.So 0.0053mol needed.


The balancing ratio, already in the reaction equation, is 1 to 2, so for 2.3 mol Zn double HCl is needed: 2*2.3 = 4.6 mol HCl


For each mole of hydrogen gas (H2) reacting with chlorine gas (Cl2), you will get 2 moles of HCl. H2 + Cl2 = 2 HCl



First, write a balanced equation for this reaction. The reactants are HCl and Zn and the products are ZnCl2(aq), and H2(g). For how to write a balanced equation, see the Related Questions to the left. Then, convert the grams of Zn into moles of Zn. To do that, see the Related Questions to the left. Then use stoichiometry to determine how many moles of HCl are necessary to react with that number of moles of Zn. See the Related Questions to the left for how to solve stoichiometry problems. Finally, determine how many milliliters of solution you need to get that many moles of HCl. To do that, use this equation: number of moles = number of liters * molarity


FeS + 2HCl >> FeCl2 + H2S 75 grams FeS (1mole FeS/87.92 grams)(2 mole HCl/1 mole FeS) = 1.71 moles HCl 2 Molar HCl = 1.71 moles HCl/Liters = 0.855 Liters HCl, or as asked for; 855 milliliters of hydrochloric acid needed


How many moles of HCl are present in 154 g of HCl?


1st Get the balanced equation NaOH + HCl -> NaCl + H2O Find the number of moles in HCl; n = cv n = 0.46x0.61 n = 0.2806 moles the number of moles of HCl and NaOH is the same so 0.2806moles will be needed


25 mL HCl x 1L/1000mL x .200 moles HCl/1L HCl=.005 moles HCl


8.3 grams HCl (1 mole HCl/36.458 grams) = 0.23 moles HCl ------------------------


Balanced equation first. NaOH + HCl >> NaCl + H2O Everything is one to one, so 1.222 moles HCl (1mole NaOH/1mole HCl) = 1.222 moles NaOH


Molarity = moles of solute/Liters of solution (50 ml = 0.05 Liters ) 10 M HCl = moles HCl/0.05 Liters = 0.5 moles HCl