And $Z$ defines the identity of the element: $Z=1,H;Z=2,He;..........Z=6,C;........Z=37, Ru...........$ You don't have to remember these numbers, because a Periodic Table will be supplied to you in every test of...
1 Answers 1 viewsSo you have the balanced equation, that's a good start. So what we're looking for is the ratio of hydrogen gas produced per the limiting reactant, let's find that first,...
1 Answers 1 viewsThe idea here is that chemical reactions must obey the Law of , which states that when a chemical reaction takes place, the total mass of the reactants must be...
1 Answers 1 viewsBecause the magnesium ion has a +2 charge, the formula for magnesium chloride is $MgCl_2$. To balance the four Cl ions in titanium (IV) chloride, the reaction must produce 2...
1 Answers 1 viewsOkay, so you've performed the classic magnesium oxide lab. I assume that you have your data, so I'll just walk you through this from a theoretical standpoint. So, you...
1 Answers 1 viewsAs do many active metals, magnesium nitride can be formed by heating the metal (fiercely!) under dinitrogen gas: $3Mg + N_2 stackrel(Delta)rarr Mg_3N_2$ As a nitride it can by hydrolyzed...
1 Answers 1 viewsThese are regular double-replacement reactions, of the form $AB + CD -> AD + CB$. Basically, the cations switch, and charge is considered when determining the number of...
1 Answers 1 views$MgCO_3(s)+DeltararrMgO(s) + CO_2(g)$ And we would have to heat the magnesium salt VERY fiercely to get complete decarboxylation. $"Moles of carbon dioxide LOST"$ $=$ $(4.40*g)/(44.01*g*mol^-1)$ $=0.010*mol$ And therefore in...
1 Answers 1 viewsThe reaction happening in solution is the following: $ZnO(aq)+2HCl(aq)->ZnCl_2(aq)+H_2O(l)$ Since the hydrochloric acid ($HCl$) is in excess, therefore, the number of moles of Zinc oxide ($ZnO$) will determine the mass...
1 Answers 1 viewsBalanced equation of the reaction is $Mg+2H_2O->Mg(OH)_2+H_2 uarr$ Molar masses of in reactants of Magnesium: 24.3050 Atomic mass of Hydrogen: 1.00794 Atomic mass of Oxygen: 15.9994 Given 12...
1 Answers 1 views