neutralization reaction between Al(OH)3 and HNO3 produces the salt with which formula?
457637
Aluminum Nitrate, Al(NO3)3, since, in a neutralization you take off any acidic hydrogen's, and any hydroxyls, and you combine the remaining ions in a way that the product molecule has no net charge. So\n Al(OH)3=Al(+3) + 3OH(-) \n HNO3 = H(+) + NO3(-)\nThe H(+) ions and OH(-) ions combine to form water, and each Al(+3) ion combines with 3 NO3(-) ions to produce Al(NO3)3 \n\nFormic acid (CHOOH) and sodium hydroxide(NaOH) produce Sodium formate, CHOONa, for the same reason as above.\n\nNote: The other H in Formic Acid is non-acidic, and so is not replaced by Na
944583
a) [O 2-]\nb) [ I - ]\nc) [Na+]\nd) [Al 3+]\ne) [Ni 2+]\nf) [Mg 2+]
175516
I not sure what style you learned to use to balance chemical equation but here is one way to do it. \nWe have four elements to balance (Al,H,Cl,O)\nLeave the compounds with aluminum with one of each. (Al done)\nThe simplest compound Al is AlCl3. To match the Cl on both sides put a 3 in front on HCl (Cl done)\nIf you add up the H (hydrogen) on the left side of the equation you'll see 6 of them. Add a 3 in front of the H20 to balance that. (H done)\nDo a quick check and you'll see there are 3 oxygen on each side (0 oxygen)\nAll the elements have been balanced, yielding a final equation of:\n3HCl + Al(OH)3 ---> 3H20 + AlCl3
625685
Aluminium has a valanly of 3 Oxgygen has 2\nSo\n4Al + 3O2   -> 2Al2O3
588293
When an acid reacts with an alkali (base), neutralisation occurs as the pH of the mixture becomes neutral, pH=7. (Acidic: pH < 7, Basic: pH > 7).\n\nSo an example would be to react sulphuric acid with sodium hydroxide to form sodium sulphate (salt) and water.\n2NaOH + H2SO4 ---> Na2SO4 + 2H20\n\nAn ELEMENT that would neutralise sulphuric acid (i.e. make its pH 7) could be any metal.\n\ne.g. Mg + H2SO4 ---> MgSO4 + H2\n\nBut this is not called a neutralisation reaction.
889216
the balanced equation of the reaction is\n 2 Al + Fe2O3 --> Al2O3 + 2 Fe\n\n1 mole of iron(iii) oxide or aluminum oxide is equivalent to two moles of aluminum.  since you used the heats of formation of ferric oxide and aluminum oxide, you have to divide it by 2.
412530
1) H3COCH3 (or C2H6O) + 3O2 (g) --> 2CO2 (g) + 3H2O (g)\n2) HNO2 (aq) + NaOH (aq) <--> NaNO2 (aq) + H2O (l)\n3) 2K+ + Cr2O7(2-) + Fe(NO3)2 + 2H+ <--> CrO4(-) + Fe3+ + 2K+ + 2NO3(-) \n4) Na3PO4 + NH4NO3 = No reaction, all soluble salts.\n5) 2HCl (aq) + Na2S (s) --> 2NaCl (aq) + H2S (g)
564023
acid + base ---> salt + H2O
806488
NaOH + H2O---> Na+   +   OH-   +H20\n\nThere is no reaction, but sodium hydroxide will ionize in the water. Actually the water will ionize also.
222097
OH-(aq) + HC2H3O2(aq)---> H2O(l) + C2H3O2-(aq)
795171
0,240 M/L*2,60 L = 0,624 moles of Al(NO3)3\nNow, every molecule of this compound has 3 moles of NO3- ions, therefore\n0,624 moles of Al(NO3)3*(3 moles NO3-/1mol Al(NO3)3)=\n1,87 moles of NO3- ions
904934
2HNO3 + Mg(OH)2 -->Mg(NO3)2 +2H20\n\n2H+(aq) + 2NO3-(aq) + Mg+(aq) + 2(OH)-(aq) -->Mg+(aq) + 2(NO3)-(aq) +2H20(l)\n\n2+(aq) + 2(OH)-(aq) --> 2H20(l)
852236
I would say Aluminum Chloride. (AlCl3)
921837
NaHCO3 --> Na+ + HCO3-\nCa(NO3)2 --> Ca(2+) + 2NO3-\nAl2(CO3)3 --> 2Al(+3) + 3CO3 (-2)
367193
Draino is essential NaOH add that to aluminum and you will get NaOH + Al --> Al(OH)3 + Na2O + H2. So I guess hydrogen gas.
802803
The answer is:\n2Al(HCO3)3  -------> Al2(CO3)3  +  3CO2  +  3H20\n\nThe equation is now fully balanced as you have\n2 moles of Al on each side\n6 moles of H on each side\n6 moles of C on each side\n18 moles of O on each side\n\nHope that helps!
985144
My guess is dilute hydrochloric acid, to make soluble FeCL3, and AlCl3.