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Balanced equation. 2Na + Cl2 >> 2NaCl 46 grams sodium = 2 mol 23 grams Chlorine = 0.65 mol ( I think Chlorine is limiting ) 0.65 mol Cl (2mol Na/1mol Cl ) = 1.3 mol ( you do not have that; Cl limits ) 0.65 mol Cl (2mol NaCl/1mol Cl2 )(58.44g/1mol NaCl ) = 75.9 grams
The sodium metal will react with chlorine gas to form sodium chloride. The reaction involves a single displacement reaction where sodium replaces chlorine in its diatomic form to form NaCl. The final mass of the flask will increase by 2.0 grams due to the formation of sodium chloride.
600 mL of 0,9 % sodium chloride: 6 x 0,9 = 5,4 grams NaCl
The molecular weight of NaCl is 58.44; sodium =22.99; Chlorine=35.45. A 1 molar solution is the molecular weight in grams in 1 litre of water, so a 3.5 molar solution would be 58.44g multiplied by 3.5, which is 204.54g in 1L.
Yes, there is a limit to the amount of sodium chloride that can dissolve in 30 grams of water. This limit is known as the solubility of sodium chloride in water. At room temperature, approximately 36 grams of sodium chloride can dissolve in 100 grams of water.
To find the answer, we multiply the 7 grams of NaCl by the ratio of the molar mass of chlorine over the molar mass of sodium chloride. By doing this, we find that there are about 4.25 grams of chlorine in 7 grams of NaCl.
Since sodium chloride has equal parts of sodium and chlorine by weight, you would need 29.3 grams of sodium to create 29.3 grams of sodium chloride.
Since the reaction involves the formation of sodium chloride (NaCl), the molar ratio of sodium to chlorine is 1:1. Therefore, if 10 grams of sodium reacts to form 35 grams of NaCl, then 35 grams of chlorine is also needed for this reaction.
You would need 35.5 grams of chlorine to combine with 23 grams of sodium to make 58.5 grams of sodium chloride. This is based on the molar mass of sodium chloride, where one mole of sodium chloride is composed of one mole of sodium (23 grams) and one mole of chlorine (35.5 grams).
75 g sodium chloride contain 29,75 g sodium.
To produce 10 grams of table salt (sodium chloride), you would need roughly 10 grams of chlorine gas. This is because the balanced chemical equation for the reaction between sodium and chlorine to form sodium chloride (table salt) is 2Na + Cl2 -> 2NaCl.
Balanced equation. 2Na + Cl2 >> 2NaCl 46 grams sodium = 2 mol 23 grams Chlorine = 0.65 mol ( I think Chlorine is limiting ) 0.65 mol Cl (2mol Na/1mol Cl ) = 1.3 mol ( you do not have that; Cl limits ) 0.65 mol Cl (2mol NaCl/1mol Cl2 )(58.44g/1mol NaCl ) = 75.9 grams
The molar mass of sodium chloride is 58.44 g/mol. To find the mass of sodium chloride formed, you need to compare the moles of sodium and chlorine to determine the limiting reactant. Calculate moles of sodium and chlorine, determine limiting reactant, and use stoichiometry to find mass of sodium chloride formed.
To calculate the total amount of sodium chloride needed for a 13 L solution at 4 grams per liter, multiply the concentration by the volume of the solution: 4 grams/L x 13 L = 52 grams of sodium chloride. Therefore, you will need 52 grams of sodium chloride to make the 13 L solution.
The answer is 1,105 g.
When sodium reacts with chlorine gas in a 1:1 molar ratio, the resulting compound is table salt (sodium chloride). The molar mass of sodium chloride is approximately 58.5 g/mol. Therefore, when you combine 1.5 grams of chlorine gas with sodium, you would produce approximately 58.5 grams of table salt.
The sodium metal will react with chlorine gas to form sodium chloride. The reaction involves a single displacement reaction where sodium replaces chlorine in its diatomic form to form NaCl. The final mass of the flask will increase by 2.0 grams due to the formation of sodium chloride.