This is a reaction, a type of double replacement reaction. The products are a salt and water. By writing the reactants on the left side as ions, you can see that the hydrogen ion,
The law of conservation of matter/mass states that matter is neither created nor destroyed during a chemical equation in a closed system. So the total mass of the reactants equals...
1 Answers 1 viewsThere would also be some heat evolved as the metal reacted. But this is no substitute for doing the experiment yourself (safely of course, while you wear safety specs to...
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 viewsMagnesium hydroxide plus hydrochloric acid produce magnesium chloride and water. $"Mg(OH)"_2("s") + "2HCl(aq)"$$rarr$$"MgCl"_2("aq") + "2H"_2"O(l)"$ You could evaporate the water to obtain solid $"MgCl"_2"$. The insoluble base is magnesium hydroxide...
1 Answers 1 viewsAnd thus we write the stoichiometric equation: $H_3PO_4(aq) + Mg(OH)_2(s) rarr Mg^(2+)HPO_4^(-)(aq) + 2H_2O(l)$. Given the 1:1 (and 1:2 stoichiometry with respect to water, $0.48*mol$ of water will be...
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 viewsI'll assume you mean what mass of aluminum hydroxide reacted. We can use the coefficients of the equation to determine the number of moles of $"Al(OH)"_3$ that react, knowing that...
1 Answers 1 viewsIndeed, your approach here will be to use the molar mass of aluminium hydroxide to convert the mass of the sample to moles and the mole ratio that exists between...
1 Answers 1 views$Mg(s) + 2HCl(aq) rarr MgCl_2(aq) + H_2(g)uarr$ If there are $80*g$ salt produced, this represents a molar quantity of $(80*g)/(95.21*g*mol^-1)=0.840*mol$. And thus there MUST have been an $0.840*mol$ quantity of...
1 Answers 1 viewsWe gots $Al(OH)_3$...even tho this is still a basic species. $"Alumina"$, $Al_2O_3$, is an amphoteric oxide...that could act as an acid. The hydroxides of lithium, calcium, or sodium, all give...
1 Answers 1 views