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The formula for aluminum sulfate is $Al_2(SO_4)_3$. What is the molar mass of aluminum sulfate to the nearest gram?
The idea here is that the molar mass of aluminium sulfate, $"Al"_ 2("SO"_4)_color(blue)(3)$, will be equal to the sum of the molar masses of all the atoms that make up...
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You are given quantities of aluminum sulfate, $Al_2(SO_4)_3$, and sulfuric acid, $H_2SO_4$. If there are EQUAL numbers of SULFUR atoms with respect to each salt, what is the ratio of oxygen atoms...$Al_2(SO_4)_3:H_2SO_4$?
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...
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Aluminum sulfate, $Al_2(SO_4)_3$, decomposes to form aluminum oxide, $Al_2O_3$, and sulfur trioxide, $SO_3$. How do you write the balanced equation for this reaction?
Putting 3 as cooefficient for sulfur trioxide will balance the chemical equation. Check the number of atoms of each element by having an inventory. Reactant Side: $Al=2xx1=2$ $S=3xx1xx1=3$ $O=3xx4xx1=12$ Product...
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How many moles of aluminum, sulfur, and oxygen atoms are there in 7.00 moles of aluminum sulfate, $Al_2(SO_4)_3$?
In one mole of aluminum sulfate, $"Al"_2"(SO"_4")"_3"$, there are 2 moles Al atoms, 3 moles sulfur atoms, and 12 moles O atoms. To determine the number of moles of atoms...
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Aluminum will react with sulfuric acid in the following reaction: $2Al(s) + 3H_2SO_4(l) -> Al_2(SO_4)_3(aq) + 3H_2(g)$. How many moles of $H_2SO_4$ will react with 18 mol Al? How many moles of each product will be produced?
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...
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When 7.25 g of $Al$ reacts with excess $H_2SO_4$ in the reaction $2Al (s) + 3 H_2SO_4 (aq) -> Al_2(SO_4)_3(aq) + 3H_2(g)$ how many liters of $H_2$ is formed at 25° C and a pressure of 1.650 atm?
$2$ $Al(s)$ + $3$ $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 $27$...
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What is the mass of copper produced if 4.87 g of $Al$ reacts with copper sulfate in the reaction $2Al + 3CuSO_4 -> Al_2(SO_4)_3 + 3Cu$?
$2Al + 3CuSO_4 rarr Al_2(SO_4)_3 + 3Cu$ $"Moles of aluminum " =(4.87*g)/(26.98*g*mol^-1)$ $=$ $0.181$ $mol$ From the equation, we know that $3$ moles of copper are reduced per $2$ mol...
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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 $Al_2(SO_4)3$?
Since aluminium sulphate is in excess, it implies that sodium hydroxide is the limiting reactant and decides how much product is formed as it gets used up first. The balanced...
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In the equation $Al_2(SO_4)_3 + 6NaOH -> 2Al(OH)_3 + 3Na_2SO_4$, what is the mole ratio of $NaOH$ to $Al(OH)_3$?
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....
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Using the balanced equation $3Ba + Al2(SO_4)_3 -> 2Al + 3BaSO_4$, if 4g Ba react with 0.45 g $Al_2(SO_4)_3$, what is the excess reactant?
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...
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