How many grams of N2F4

  1. 14.0g of F2 was reacted with excess NH3 to produce N2F4 and HF. The equation for the reaction: 5F2(g) + 2NH3(g) - N2F4(g) +
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  2. 5F2(g) + 2NH3(g) ==> N2F4(g) + 6HF(g)a)If you have 66.6 g NH3, how many grams of F2 are required for complete reaction? b)How
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    2. Danny asked by Danny
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  3. what do limiting reagents have to do with...5F2(g) + 2NH3(g) ==> N2F4(g) + 6HF(g) a)If you have 66.6 g NH3, how many grams of F2
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    2. Keith asked by Keith
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  4. 5F2(g) + 2NH3(g) ==> N2F4(g) + 6HF(g) If you have 225 g F2, how many grams of N2F4 will be produced
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    2. Kyra asked by Kyra
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  5. Based on the balanced equation,10Li + N2F4 → 4LiF + 2Li3N calculate the formula units of LiF formed when 54 molecules of N2F4
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    2. eng asked by eng
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  6. How many grams of N2F4 can be produced from 361 g F2?
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    2. Marlon asked by Marlon
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  7. if 68.0g of NH3 react with 304g of F2, how many grams of N2F4 are produced
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    2. annaleisa asked by annaleisa
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  8. Need help with my last question!Given the following reaction: 2NH3 (g) + 5F2 (g) ---> N2F4 (g) + 6HF (g) If 4.00 g of NH3 and
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    2. kaitie asked by kaitie
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  9. The reaction of flourine with ammonia produces dinitrogen tetrafluoride and hydrogen fluoride.5F2(g) + 2NH3(g) ==> N2F4(g) +
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    2. Keith asked by Keith
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  10. The reaction of fluorine with ammonia produces dinitrogen tetra fluoride and hydrogen fluorida.5F2(g)+2NH(g)---> N2F4(g)+6HF(g)
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    2. Lexa asked by Lexa
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