How to solve this chemical equation??? chemistry?
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I dunno how to solve this equation please help.. 2Ag2*CO3* = 4Ag + 2CO2* + O2* * sign is subscript Find the mass of silver carbonate, Ag2*CO3* that is required to produce 10 g of silver. the answer is actually 12.78 g but every time i tried to solve the equation, i always get 13.8 g.. PLEASE HELP ME!! ( i'll be punish by my teacher if i can't do this )
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Answer:
Atomic mass Ag = 107.87, formula mass Ag2CO3 = 275.74 1 mole Ag2CO3 produces 2 moles Ag therefore 275.74g Ag2CO3 produces 215.74 g Ag so 10 g Ag requires (275.74/215.74) x 10 = 12.78 g Ag2CO3
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Other answers
Atomic mass Ag = 107.87, formula mass Ag2CO3 = 275.74 1 mole Ag2CO3 produces 2 moles Ag therefore 275.74g Ag2CO3 produces 215.74 g Ag so 10 g Ag requires (275.74/215.74) x 10 = 12.78 g Ag2CO3
sandie
From the reaction: 2 moles (551.48g) of Ag2CO3 produce 4 moles (431.48g) of silver metal. Hence 10g of silver metal could be obtained from (551.48 x 10)/431.48 = 12.7811g of Ag2CO3.
Umar
whats wrong with the teacher, thats quite close answer. ignore ur teacher
pepsi_coke
From the reaction: 2 moles (551.48g) of Ag2CO3 produce 4 moles (431.48g) of silver metal. Hence 10g of silver metal could be obtained from (551.48 x 10)/431.48 = 12.7811g of Ag2CO3.
Umar
whats wrong with the teacher, thats quite close answer. ignore ur teacher
pepsi_coke
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