In the reaction $Al_2(SO_4)_3 + 6NaOH -> 2Al(OH)_3 + 3Na_2(SO_4)_3$, how many moles of $Al(OH)_3$ can be made with 2.3 moles of $NaOH$ and excess $Al_2(SO_4)3$?

It is given that both salts contain the same number of sulfur atoms. Thus we have $Al_2S_3O_12$, and necessarily $3$ $xx$ $H_2SO_4$. There are $3$ $"equiv"$ sulfuric acid per equiv...

$2NaOH(s) + CO_2(g) -> Na_2CO_3(s) + H_2O(l)$ Molar mass of $NaOH = 40 $ $2xx40 g$ of $NaOH$ produces $6.02 xx 10^23$ molecules of $Na_2CO_3$ $:.$ $10 g$ of $NaOH$...

With reference to the given balanced equation for the , the needed number of moles for each remaining compound involved in the reaction using the mass of $Al$ as basis...

$2$ $Al(s)$ + $3$ $H_2SO_4(aq) to $ $Al_2(SO_4)_3(aq)$ + $3$ $H_2(g)$ First calculate the number of moles of $Al$ in $7.25$ $g$, knowing the molar mass of $Al$ is $27$...

$2Al + 3CuSO_4 rarr Al_2(SO_4)_3 + 3Cu$ $"Moles of aluminum " =(4.87*g)/(26.98*g*mol^-1)$ $=$ $0.181$ $mol$ From the equation, we know that $3$ moles of copper are reduced per $2$ mol...

Given the stoichiometric equation, you have established the . $"Moles of dihydrogen"$ $=$ $(9.9*g)/(2.0*g*mol^-1)$ $=$ $??*mol$. The moles of aluminum metal is $2/3$ this quantity. And of course $"1...

The balanced equation will tell the stoichiometric ratio of reactants and products. Here in the balanced equation the coefficient before $NaOH$ is 6 and that before $Al(OH)_3$ is 2....

Balanced Equation $"3Ba + Al"_2"(SO"_4)_3"$$rarr$$"2Al + 3BaSO"_4$ The coefficients are the moles of each substance, and they can be used as molar ratios between two of the substances. No coefficient...

For any chemical reaction, the balanced chemical equation tells you the ratio that must always exist between the reactants. In your case, you have $"FeCl"_ (2(aq)) + color(blue)(2)"NaOH"_...

The idea here is that you're diluting the stock solution to a total volume of $overbrace("5 mL")^(color(blue)("the volume of the stock solution")) + overbrace("300 mL")^(color(blue)("the volume of water"))...