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What is Hcl g name?

Updated: 9/27/2023
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Q: What is Hcl g name?
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What is he necessary amount of chlorine needed to prepare 25.00 g of hydrogen chloride by reacting hydrogen with chlorine?

H2 (g) + Cl2 (g) --> 2 HCl (g) 25.00 g HCl x 1 mol HCl x 1 mol Cl2 x 70.90 g Cl2 = 24.3 g Cl2 are needed. ................... 36.46 g HCl . 2 mol HCl .. 1 mol Cl2


if 6.5g of zn reacts with 5g of hcl?

7.3 g of HCl.


What is an explaination of 0.01 M hydrochloric acid?

1M Hcl contains 36.5 g of HCl per liter of water. 0.01M HCl thus contains 0.365 g of HCl per liter.


How many moles HCl in 8.3 g HCl?

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


Why standard normality of HCl is 11.3 N?

g HCl solution = 2500 mL of HCl * 1 liter/1000 mL * 1190 g/L = 2975 g 37% solution (37 g HCl/100 grams of solution) gives you the grams of HCl: g HCl = 0.37 * 2975 g = 1100.8 g HCl Moles HCl = 1100.8/(36.46 g/mole) = 30.2 moles Therefore the molarity, which equals the normality in this case = 30.2 moles/2.5 L = 12.07 M = 12.07 N If you want to make 100 mL of a 0.1 N solution, Volume of HCl solution needed = (0.1 N * 100 mL) /12.07 N = 0.83 mL Take 0.83 mL of the 37% HCl, and dilute it with water to 100 mL.


How many grams of hydrogen is produced when 167.12 g of hydrochloric acid is reacted with copper?

Cu + 2 HCl --> CuCl2 + H2 167.12 g HCl * (1 mol HCl/36.46 g HCl) * (1 mol H2/2 mol HCl) * (2.02 g/1 mol H2) = 4.629 g H2 4.63 grams of hydrogen gas is produced when 167.12 grams of hydrochloric acid reacts with copper.


How many liters of HCl are produced by the reaction of 5.7 L of hydrogen with an equal amount of chlorine H2 g Cl2 g -- 2 HCl g?

11.45


What mass of hydrogen chloride is require to react with 100 g of rust?

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


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

Divide by molar mass and check the units(italicalized):0.140 (g HCl) / 36.45 (g.mol-1HCl) = 3.84*10-3 mol HCl


Find out the mass of HCL needed to react completely with 0.20 g of ammonia gass?

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


What is the molarity of 32 percent HCl?

A: 10.172 M HCl. Percents of acids like this one is normally given in % w/w, or percent weight by weight. So we need to find the mass of the solution. Molarity is moles of solute per liter solution, so we will calculate this for 1L of solution. The density of 32% HCl is 1.159 g/mL. So assuming we have one liter or 1000 mL, the weight of the solution is 1159 g. If 32% of this is HCl, we have 1159 g solution * 0.32 g HCl/g solution = 370.88 g HCl. The molar mass of HCl is 36.461 g/mol, so 370.88 g HCl * 1 mol HCl/36.461 g HCl = 10.172 mol HCl in 1 L of solution. 10.172 mol HCl/1L solution = 10.172 M HCl.


How many grams of HCl are contained in 4.25mL of a 0.405M aqueous solution of HCl?

M * V = n 0.405 M * 0.00425 ml = 0.00172125 mole of HCl The molar mass of HCl is: 1.007947 + 35.453 = 36.460947 g/mole m = mm * n So the mass in gram is: 36.460947 g/mole * 0.00172125 mole = 0.0628 gram