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How many atoms of zinc react with 1.49 g HNO3

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Start with this formula: 4 Zn(s) + 10 HNO3 –>4 Zn(NO3)2 + NH4NO3 + 3 H2O then divide 1.49 by the molar mass. That’s a start! [ Source: http://www.chacha.com/question/how-many-atoms-of-zinc-react-with-1.49-g-hno3 ]
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How many atoms of zinc react with 1.49 g HNO3
http://www.chacha.com/question/how-many-atoms-of-zinc-react-with-1.49-g-hno3
Start with this formula: 4 Zn(s) + 10 HNO3 –>4 Zn(NO3)2 + NH4NO3 + 3 H2O then divide 1.49 by the molar mass. That’s a start!

Related Questions Answered on Y!Answers

Last stoichiometry problem I need help with, Please!?
Q: I would be GRATEFUL if someone could help me by work just one of my problems on my work sheet so I would hopefully know what direction to go in with the rest of them. Please work through it in some detail, I am not just looking for an answer, in fact I have the answer, I only wish to understand how to work these problems, heres the problem:4Zn + 10HNO3 —–> 4Zn(NO3)2 + NH4NO3 + 3H2Oa. How many atoms of zinc react with 1.49 g HNO3?b.Calculate the number of grams of zinc that must react with an excess of HNO3 to from 29.1 g NH4NO3
A: Moles HNO3 = 1.49 g / 63.01 g/mol =0.0236the ratio between Zn and HNO3 is 4 : 10moles Zn needed = 0.0236 x 4 / 10 = 0.00946atoms Zn = 6.02 x 10^23 x 0.00946 = 5.69 x 10^21Moles NH4NO3 = 29.1 g/ 80.045 g/mol = 0.364the ratio is 4 : 1Moles Zn = 0.364 x 4 = 1.46grams Zn = 1.46 mol x 65.39 g/mol =95.2 g
PLEASE HELP! CHEMISTRY HELP!?
Q: PLEASE ANSWER ANYTING YOU CAN! :)1. Nitric acid and zinc metal react to form zinc nitrate, ammonium nitrate, and water.HNO3(aq) + Zn(s) –> Zn(NO3)2(aq) + NH4NO3(aq) + H2O(l)a) Balance this reaction.b) How many atoms of zinc react with 1.49 grams of HNO3?c) Calculate the number of grams of zinc that must react with an excess of nitric acid to form 29.1 grams of ammonium nitrate.2. If the reaction below proceeds with a 96.8% yield, how many kilograms of CaSO4 are formed when 5.24 kg of SO2 reacts with an excess of CaCO3 and O2?CaCO3(s) + SO2(g) + O2(g) –> CaSO4(s) + CO2(g)Thank you!
A: 1. Nitric acid and zinc metal react to form zinc nitrate, ammonium nitrate, and water.HNO3(aq) + Zn(s) –> Zn(NO3)2(aq) + NH4NO3(aq) + H2O(l)a) Balance this reaction.Zn loses 2 e-1 HNO3 takes 8 e- –> 1 NH4+we need 4 Zn’s to balance electrons& 9 extra HNO3’s to supply the (NO3)’s10 HNO3(aq) + 4 Zn(s) –> 4 Zn(NO3)2(aq) + 1 NH4NO3(aq) + 3 H2O(l)====================================================b) How many atoms of zinc react with 1.49 grams of HNO3?1.49 g Zn @ (10 mol HNO3 @ 63.0 g/mol) / 4 mol Zn @ 65.4 g/mol)= your answer is 3.59 grams of HNO3===================================================c) Calculate the number of grams of zinc that must react with an excess of nitric acid to form 29.1 grams of ammonium nitrate.29.1 g NH4NO3 @ (4 mol Zn @ 65.4 g/mol) / (1 mol NH4NO3 @ 80.04 g/mol) =your answer is95.1 g of NH4NO3==========================================2. If the reaction below proceeds with a 96.8% yield, how many kilograms of CaSO4 are formed when 5.24 kg of SO2 reacts with an excess of CaCO3 and O2?CaCO3(s) + SO2(g) + O2(g) –> CaSO4(s) + CO2(g)find theoretical yield:5.24 kg SO2 @ 136.14 g/mol CaSO4 / 64.0 g/mol SO2 =11.15 g CaSO4 could be [email protected] 96.8% of 11.15 g CaSO4 could be produced;10.8 grams of CaSO4 are produced
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