This equation tells us that $26.98*g$ aluminum metal gives us $33.6*L$ of dihydrogen gas. Why? Because $1$$mol$ of gas occupies $22.4*L$ at $"STP"$ and $1.5*mol$ were generated.
$"Moles of aluminum"$$=$$(30.2*g)/(26.98*g*mol^-1)$$=$$1.12*mol$.
Given the , $3/2xx1.12*cancel(mol)xx22.4*L*cancel(mol^-1)$ are generated.
We calculate the relative quantities from the molar ratios in the balanced reaction equation. $2CO(g)+O_2(g)→2CO_2(g)$. As a pretty good approximation, gas volumes are proportional to molar amounts. Therefore with...
You know that $2"Na"_ ((s)) + "Cl"_ (2(g)) -> 2"NaCl"_ ((s))$ and that the reaction produced $"234 g"$ of sodium chloride. Convert this to moles by using the...
We're asked to find the number of grams of $"Al"$ that reacted, given some $"H"_2 (g)$ product characteristics. Let's first write the chemical equation for this reaction: $2"Al"(s) +...
Each equiv of metal reduces $3$ equiv of acid, and $3/2$ equiv of dihydrogen gas are evolved. $"Moles of aluminum"$ $=$ $(0.498*g)/(26.98*g*mol^-1)=0.0185*mol$. Now, given the conditions, clearly the metal is...
Given: Volume of $"H"_2$ at STP; $P$ and $T$ of $"HCl"$; chemical equation (understood) Find: Volume of $"HCl"$ Strategy: The central part of any problem is to convert moles...
The answer is $"21.5 L" $. So, start with the balanced chemical equation for the decomposition of hydrochloric acid $2"HCl" -> "H"_2 + "Cl"_2$ Notice that you have a $"2:1"$...
The thing to remember about reactions that involve gases held under the same conditions for pressure and temperature is that the mole ratio that exists between the gaseous in the...
$2.5$ $liters$ of a $0.08$ $molar$ solution of $HCI$ gives us $mol$$es$ of $HCl$ in the solution $Molarity$*$Volume$ = $mol$$es$ $0.08$ $(mol)/cancel L$ * $2.5 cancel L$ = $0.2 mol$$es$...
$"Moles of HCl"$ $=$ $6.00*Lxx0.100*mol*L^-1=0.600*mol$. So we need a volume of conc. acid that contains $0.600*mol$. $"Volume"$ $="Amount in moles"/"Concentration"$ $=$ $(0.600*mol)/(12.0*mol*L^-1)xx1000*mL*L^-1$ $=$ $50*mL.$
