How much AgCl in grams can be produced from 987.65g of AgNo3???
NaCl+AgNO3--->AgCl+NaNO3 thats the equation
please help i dont get it
2007-11-29 12:12:38 · 2 answers · asked by $$$$$ 2 in Education & Reference ➔ Homework Help
Ag has an atomic weight of 107.868 and Na a weight of 22.990, so your 987.65 g of AgNa3 is 107.868 / 176.838 = 61% silver. Thus the silver content is 604.00 grams.
Cl has a weight of 35.453 so AgCl is 107.868 / 143.321 = 75% silver. You have 604.00 grams of Ag, so the total weight of the compount must be 802.52 grams.
2007-11-29 12:26:52 · answer #1 · answered by jgoulden 7 · 0⤊ 0⤋
i got 833.6 g of AgCl
first, divide the amount of AgNO3 by its mass, 169.9 g. in order to get the number of AgNO3 moles present. you should get about 987.65/169.9=5.813 moles of AgNO3.
then, you should multiply by the molar ratios. since the molar ratio is one in this case, you don't really need to do it. you should now have 5.813 moles of AgCl
finally, multiply 5.813 moles of AgCl by its mass, 143.4 g in order to get the mass of AgCl produced.
you should get 833.5842 g of AgCl, or 833.6 g of AgCl if you need the correct number of significant figures.
i have a test over this tomorrow, so if that isn't right, then i'm screwed too... lol
2007-11-29 20:43:22 · answer #2 · answered by tired 1 · 0⤊ 0⤋