Of course, we could do this analytically, and write the chemical equation for water decomposition, and then apportion molar quantities: $H_2O(l) rarr H_2(g) + 1/2O_2(g)$, but why should I bother?...
1 Answers 1 viewsI'll show you two approaches to solving this problem, one really short and one relatively long. $color(white)(.)$ SHORT VERSION The problem tells you that $"6 g"$ of hydrogen gas,...
1 Answers 1 viewsSimply write out ratio of the known and unknown components: $(n(NH_3))/(n(N_2))=2/1$ $n(NH_3)=2n(N_2)=2xx1.75mol=3.5mol$
1 Answers 1 viewsThe stoichiometrically balanced equation tells us unequivocally that $1$ $mol$ dihydrogen reacts with $0.50$ $mol$ dioxygen to give $1$ $mol$ water. $"Moles of dihydrogen"$ $=$ $(58.1*g)/(2.02*g*mol)$ $=$ $28.7*mol$ $H_2$ $"Moles...
1 Answers 1 viewsWe need a stoichiometric equation for water synthesis: $H_2(g) + 1/2O_2(g) rarr H_2O(l)$ Clearly, dihydrogen must be present in a 2:1 molar ratio with respect to dioxygen. $"Moles of dihydrogen"$...
1 Answers 1 viewsWith 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...
1 Answers 1 viewsFrom the equation above, 3 moles of carbon dioxide result from every mole of propane combusted. We had, $(7.51*g)/(44.1*g*mol^-1)$ $=$ $?? "moles of propane"$. So how many moles of...
1 Answers 1 viewsStart by balancing the equation for the reaction between hydrogen $"H"_2$ and nitrogen $"N"_2$, which produces ammonia $"NH"_3$: $color(darkgreen)(1) color(white)(.) "N"_2 (g) + color(darkgreen)(3) color(white)(.) "H"_2 to color(darkgreen)(2) color(white)(.) "NH"_3...
1 Answers 1 viewsEach 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...
1 Answers 1 viewsThe first thing that you need to do here is to use the equation $color(blue)(ul(color(black)(PV = nRT)))$ Here $P$ is the pressure of the...
1 Answers 1 views