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How much hcl gas can be produced at stp by reacting 131 g of nacl with excess h2so4?
Using the equation: 2 NaCl + H2SO4 -> 2 HCl + Na2SO4, we can see that 1 mole of NaCl will produce 1 mole of HCl. First, calculate the moles of NaCl (131g / 58.44g/mol). Then, using the mole ratio from the equation, you can find the moles of HCl produced. Finally, using the ideal gas law, you can convert the moles of HCl to volume at STP.
How much NaCl precipitate would be produced if 2 molarity HCl and 1 molarity sodium carbonate react in the reaction Na2CO3 plus 2HCL yields 2NaCl plus H2O plus CO2?
2 moles or 117 gram of NaCl is precepitated
How much of NaCl is in 1.31 L of 0.300 M NaCl?
1.31*.3=.393 mols NaCl. This also equals .393(22.99+35.45)=22.97 g NaCl.
How much Elemental chloride in NaCl?
The formula unit of sodium chloride (NaCl) contain 60,33 % chlorine.
How much iodine was put in NaCl?
The recommended concentration is 20 +/- 5 mg iodine/kg NaCl.
How much gNacl and how much gwater needed for ready of 500g solution of NaCl 0.95 percent?
0.95% * 500 g = 4.75 g NaCl
Determine how much of each of the following NaCl solutions in grams contains 1.5g of NaCl 0.058 percent NaCl by mass?
In order to determine this, it is necessary to know what solution we are looking at. One we know that we can look at the grams in a mole of the substance and determine the percentages based on molecular weight.
How much nacl is required for a 200 mM NaCl solution?
You need 58,44 mg of ultrapure NaCl; dissolve in demineralized water, at 20 0C, in a thermostat, using a class A volumetric flask of 1 L.
The concertration of a water solution of NaCl is 2.48m and it contains 806 g of water How much NaCl is in the solution?
To determine the amount of NaCl in the solution, you first need to calculate the moles of NaCl present. Using the given molarity (2.48 M) and the volume of the solution (assumed to be 806 g = 806 ml for water), you can find the moles of NaCl. Then, you convert the moles of NaCl to grams using the molar mass of NaCl (58.44 g/mol) to find the amount of NaCl in the solution.
How much of NaCl is in 1.84 L of 0.200 M NaCl Answer in units of mol.?
To find the amount of NaCl in 1.84 L of 0.200 M NaCl, you use the formula: amount = concentration x volume. Thus, amount = 0.200 mol/L x 1.84 L = 0.368 mol of NaCl.
How much NaCl precipitate would be produced if 2 molarity HCl and 1 molarity sodium carbonate react in the reaction Na2CO3 plus 2HCL yeilds 2NaCl plus H2O plus CO2?
The resulting product would be a roughly 2 M NaCl solution (slightly less than 2 molar because the solution is diluted by the water that is produced by the reaction). 2 M HCl + 1 M Na2CO3 --> 2 M NaCl + 1 M H2O + 1 M CO2 Two moles of NaCl weigh about 117 g (58.5 grams per mol) so the resulting solution has a sodium chloride concentration of about 117 grams per liter. This concentration is well below the maximum solubility of water at room temperature and atmospheric pressure ( > 300 g/liter) so there would not be any NaCl at all precipitating. The answer is thus: no NaCl precipitate will form.
How much grams of chloride in 58 grams of NaCl?
58 g NaCl = 58 (g) / 58.44 (g/mol NaCl) = 0.9925 (mol NaCl) = 0.9925 (mol Cl-) = 0.9925 * 35.45 (g/mol Cl-) = 35.2 g Cl-
