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Maria_rrd
A really tough bot, or a member.
Joined: 17 Sep 2005
Posts: 2
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 Posted: Sat Sep 17, 2005 9:42 pm Post subject: Help |
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| 0.684 g of silver nitrate and .473g of potassium bromate are added to 231 mL water. Solid silver bromate is formed, dried, and weighed. What is the mass, in g, of the precipitated silver bromated? Assume silver bromate is completely insoluble.
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charco
Regular
Joined: 07 Mar 2005
Posts: 96
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| Posted: Sun Sep 18, 2005 9:49 pm Post subject: |
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1. work out the moles of each
2. find the limiting reagent
3. work out the moles of product
4. calculate the mass of product
5. retire with smug grin an large can of castlemaine 4X
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hollywood1
Probably a bot.
Joined: 24 Sep 2005
Posts: 1
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| Posted: Sat Sep 24, 2005 10:39 pm Post subject: The Equation for ur question |
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I got the same question as you just that I have 0.684 potassium bromate and 0.562 g of silver nitrate which are added to 218 ml of water. Is this the correct balanced equation?
KBrO3 + AgNO3 = AgBrO3 + KNO3
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