1M Hcl contains 36.5 g of HCl per liter of water. 0.01M HCl thus contains 0.365 g of HCl per liter.
8.3 grams HCl (1 mole HCl/36.458 grams) = 0.23 moles HCl ------------------------
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.
11.45
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
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
7.3 g of HCl.
1M Hcl contains 36.5 g of HCl per liter of water. 0.01M HCl thus contains 0.365 g of HCl per liter.
8.3 grams HCl (1 mole HCl/36.458 grams) = 0.23 moles HCl ------------------------
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.
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.
11.45
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
Divide by molar mass and check the units(italicalized):0.140 (g HCl) / 36.45 (g.mol-1HCl) = 3.84*10-3 mol HCl
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
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.
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