$BaCl_2(aq) + Na_2SO_4(aq) rarr BaSO_4(s)darr + 2NaCl(aq)$ $"Moles of barium chloride"=(18.48*g)/(208.23*g*mol^-1)=8.87xx10^-2*mol$ Barium chloride is the . At MOST, we can get $8.87xx10^-2*mol$ $BaSO_4(s)$, i.e. $8.87xx10^-2*molxx233.38*g*mol^-1=??g$ $"Barium sulfate"$ is as soluble...
1 Answers 1 viewsIn $"MgSO"_4$, the number of oxygen atoms is represented by the subscript directly next to the $"O"$. In $"7H"_2"O"$, the coefficient $7$ can be multiplied throughout each element in the...
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 viewsBalanced Equation $"Pb(SO"_4)_2("aq") + "4LiNO"_3("aq")"$$rarr$$"Pb(NO"_3)_4("aq")" + 2Li"_2"(SO"_4)("aq")"$ This at first glance looks like a double replacement (double displacement) reaction. However, a double replacement reaction does not occur unless one of...
1 Answers 1 views$NiCl_2(aq) + 2NaOH(aq) rarr Ni(OH)_2(s)darr + 2NaCl(aq)$ There are $0.0108$ $mol$ $NiCl_2$ ($22.0xx10^(-3)cancelLxx0.490*mol*cancel(L^-1)$). So, by $0.0216$ $mol$ $NaOH$ are required. There is $0.200$ $mol*L^-1$ $NaOH$ solution available, so... $(0.0216*cancel(mol))/(0.200*cancel(mol)*cancel(L^-1))xx1000*mL*cancel(L^-1)$...
1 Answers 1 viewsWe need (i) a stoichiometric equation: $Mg(s) + O_2(g) rarr MgO(s)$, and (ii), equivalent quantities of the product $MgO$: $"Moles of magnesium oxide"=(25*g)/(40.30*g*mol^-1)=0.620*mol$. Now from the of the equation, if...
1 Answers 1 viewsThis appears to be a limiting reactant problem. 1. Write the balanced equation $"2NaCN" + "H"_2"SO"_4 → "Na"_2"SO"_4 + "2HCN"$ 2. Calculate the moles of $"Na"_2"SO"_4$ formed from...
1 Answers 1 viewsYou specify a $0.5*mol*L^-1$ solution of $MgSO_4(aq)$. Since $"Concentration"$ $=$ $"Moles of solute"/"Volume of solution"$, $"Moles of solute"="Concentration"xx"Volume of solution"$. If we have a $1000*mL$ volume, then....... $"Moles of...
1 Answers 1 viewsa solubility of $35.1*g$ of the anhydrous salt in water at $20$ $""^@C$. The anhydrous salt can bind several water molecules, and so is extensively used to dry ethereal...
1 Answers 1 views(a) Calcium sulfate is sparingly soluble. In a saturated solution the aqueous ions are in equilibrium with the insoluble salt: $sf(CaSO_(4(s))rightleftharpoonsCa_((aq))^(2+)+SO_(4(aq))^(2-))$ For which $sf(K_(sp)=[Ca_((aq))^(2+)][SO_(4(aq))^(2-)]=2.4xx10^(-5)color(white)(x)"mol"^2."l"^-2)$ The solubilty $sf(s)$ of the salt...
1 Answers 1 views