As for all these problems, we first need a stoichiometrically balanced equation,
And thus between the oxide and the starting metal, there is clearly
$3"Zn" + 2"H"_3"PO"_4 -> "Zn"_3("PO"_4)_2 + 3"H"_2$ The molar ratio of zinc: hydrogen is $3:3$ so moles of zinc = moles of hydrogen. $"m"="nM"=30xx1=30$ $"g"$
1 Answers 1 views$4As$ + $ 3O_2$ -----> $2As_2O_3$ According to equation $3$ moles of oxygen gas is required to form arsenic trioxide = $2$ moles $3.4$ moles of O2 will be required...
1 Answers 1 viewsLet's check, $50*gxx20%(m_"ZnO"/m_"cream")$ $=$ $10*g$ $ZnO$. Do you know what the major use of this cream is?
1 Answers 1 viewsThe reaction happening in solution is the following: $ZnO(aq)+2HCl(aq)->ZnCl_2(aq)+H_2O(l)$ Since the hydrochloric acid ($HCl$) is in excess, therefore, the number of moles of Zinc oxide ($ZnO$) will determine the mass...
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 views$"Moles of ferric oxide "=(0.18*g)/(159.69*g*mol^-1)$ $=$ $1.13xx10^-3*mol" metal oxide"$. $"Moles of carbon monoxide "=(0.11*g)/(28.0*g*mol^-1)$ $=$ $3.93xx10^-3*mol" CO"$. $Fe_2O_3$ is the reagent in deficiency (why?), and thus $2xx1.13xx10^-3*molxx55.85*g*mol^-1~=0.150*g$ iron metal are...
1 Answers 1 views$"ZnSO"_4 + "Li"_2"CO"_3"$$rarr$$"ZnCO"_3 + "Li"_2"SO"_4$ The process used to answer this question will be: $color(red)("given mass Li"_2"CO"_3"$$rarr$$color(blue)("mol Li"_2"CO"_3"$$rarr$$color(green)("mol ZnCO"_3"$$rarr$$color(purple)("mass ZnCO"_3"$ The molar masses of lithium carbonate and zinc sulfate is...
1 Answers 1 viewsGiven: Mass of $"O"_2$; chemical equation (understood) Find: Moles of $"CO"_2$ Strategy: The central part of any problem is to convert moles of something to moles of something else....
1 Answers 1 viewsThis means that for $6mol$ of $FeO$, only $3mol O_2$ is produced (half the moles). The volume of the oxygen is dependant on temperature and pressure. E.g. : If...
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