Q Silver chloride is formed according to the reaction:Īg+(aq) + CL-(aq) = AgCl(s). Let's start again and this is my last post on this problem Convert mLto Litres (by dividing by 1000) and divide the number of moles NaCl left by this volume to give the molarity of NaCl at the end So for 0.0125 moles NaCl you would need 0.0125 m0l AgNO3 but you only have 0.00589 mol AgNO3 - this is not enough and is limiting.Īlternatively you can say that 0.00589 mol AgNO3 will react with 0.00589 mol NaCl and you have more NaCl than this (0.0125 moles) so only part of the NaCl react and the rest is left as excessĪgNO3 is limiting and determine the mass of AgCl madeĠ.00589 mol makes (166.91/1) x 0.00589 g of AgClĪs for the excess NaCl, subtract 0.00589 moles from 0.0125 moles to give the number of moles of NaCl left.

Moles AgNO3 = 0.00589 mol AgNO3 (your calculation) of moles of NaCl = 0.05 x 0.250 = 0.0125 moles of NaCl Work must be shownĬalculate the mass/volume percent of a NaCl solution in which 182g of NaCl is dissolved in enough water to give a total volume of 4.78L. (Use the mass of sodium carbonate reactant used as the starting point, along with the relevant mole ratio from the balanced equation to perform this calculation).

How many grams of NaCl would be required to react with 132 mL of 0.719 M AgNO3 solution? Answer in units of grams.Ĭalculate the theoretical yield of NaCl (using mass-to-mass calculation).

Silver nitrate (AgNO3) reacts with sodium chloride as indicated by the equation AgNO3 + NaCl → AgCl + NaNO3. can you please explain the universal steps for me What mass of solid AgBr is produced when 100.0 mL of 0.150 M AgNO3 is added to 20.0mL of 1.00 MNaBr? i keep getting the answer 3.75 g however when i checked in the back of the book it gives me 2.82 g. After the precipitate is centrifuged down, the remaining solution (called the supernatant) is tested for the concentration of Ag+ ion. However, from the standpoint of Na,ġ0.0 mL of 0.05 M AgNO3 and 10.0 mL 0.05 M NaCl are combined in a test tube and a precipitate of AgCl forms. What is the theoretical mole ratio- NaCl to NaHCO3 How do you determine the percentage error of an experimental mole ratio It would be easier for us to help you if we had the entire problem instead of just parts of it. What mass of solid aluminum hydroxide can be produced when 50.0 mL of 0.200 M aluminum nitrate is added to 200.0 mL of 0.100M potassium hydroxide? Convert grams to moles of NaCl 0.630 * 1 molĪ ball of mass 8kg falls from rest from a height of 100m.neglecting air resistance,calculate its total energy after falling a distance of 40m mass of NaCl= 0.630g mass of H20= 525g molar mass of NaCl= 58.44 This is what i did it keeps telling me the answer is wrong. help please 25g/58.44g NaCl=.4 gramsĪ mass of 0.630 g of NaCl is dissolved in 525 g of water. calculate the mass percent, molarity, molality, and mole fraction of NaCl. Assume that density of the solution is 1.0 g/mL. A solution is made by dissolving 25 g NaCl in enough water to make 1.0 L solution.

Agno3 molar mass