We need a stoichiometric equation that represents the reaction between $Mg(OH)_2$ and $HCl$, and here it is............ $Mg(OH)_2(s) + 2HCl(aq) rarr MgCl_2(aq) + 2H_2O(l)$ Here, $1$ $"equiv"$ of magnesium hydroxide...
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 viewsI am assuming that $6.0 M HCl$ is a typo, and that it should be $6.0 molL^-1 HCl$, since that makes sense in the equation. First we have to find...
1 Answers 1 views$CuCO_3(s) + 2HCl(aq)rarrCuCl_2(aq) + CO_2(g)uarr + H_2O(l)$ Or........... $CuO(s) + 2HCl(aq)rarrCuCl_2(aq) + H_2O(l)$ For each acid base reaction the product is the beautiful blue-coloured $[Cu(OH_2)_6]^(2+)$ ion, which is commonly represented...
1 Answers 1 viewsIn the chemical equation given to you $"Mg"_ ((s)) + 2"HCl"_ ((aq)) -> "MgCl"_ (2(aq)) + "H"_ (2(g))$ magnesium is a reactant, which is why you see...
1 Answers 1 viewsMoles of metal, $=$ $(4.86*g)/(24.305*g*mol^-1)$ $=$ $0.200$ $mol$. Moles of $HCl$ $=$ $100 *cm^-3xx2.00*mol*dm^-3$ $=$ $0.200$ $mol$ Clearly, the acid is in deficiency ; i.e. it is the , because...
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 viewsMoles of magnesium: $(50.0*g)/(24.31*g*mol^-1)$ $=$ $2.06$ $mol$ Moles of hydrogen chloride gas: $(75.0*g)/(36.2*g*mol^-1)$ $=$ $2.07$ $mol$ Given equimolar amounts, clearly, the gas is in deficiency as the stoichiometric equation requires...
1 Answers 1 viewsas per given balanced equation one mole Zn produces 1mole Hydrogen when reacted with excess acid. So 3 moles Zn will give 3moles Hydrogen
1 Answers 1 views$1 "mol"$ of $"Zn"$ has mass of $65.39 "g"$. The amount of $"Zn"$ is $frac{2.5 "g"}{65.39 "g/mol"} = 0.038 "mol"$. The amount of $"H"_2$ produced is the same as the...
1 Answers 1 views