Start with the equation for the reaction: $2 "LiOH"+"CO"_2 \rightarrow "Li"_2"CO"_3+"H"_2"O"$ So one mole of lithium hydroxide reacts with one-half mole of carbon dioxide. By adding up atomic masses we...
1 Answers 1 viewsWeight of $Ca$ = $40$ Weight of $Cl$ = $35.5$ Molecular formula is $CaCl_2$ i.e. $1$ atom of $Ca$ and $2$ atoms of $Cl$. Hence, molecular weight = $"1...
1 Answers 1 viewsSince the given information is in moles, we need only look at the ratio between the chemicals: Six moles of Li will produce 2 moles of lithium nitride. So, the...
1 Answers 1 viewsRubidium, an alkali metal, commonly forms $Rb^+$ salts, just as lithium commonly forms $Li^+$ salts. We would assume that the formula of rubidium nitride (if it exists!) is $Rb_3N$.
1 Answers 1 viewsAcquiring the percentage requires the usage of "unified" units, meaning units that are equivalent across the board, because a percentage is really a unitless fraction (all the involved units must...
1 Answers 1 viewsFirst step, find the molar mass of $Li_2CO_3$. The molar mass is same as atomic weight on . But you have to see what are present in this formula. You...
1 Answers 1 viewsSo we have the $"%calcium carbonate, "m/m$. We are not finished yet, because we have to address the percentage by mass of calcium metal in calcium carbonate: $%Ca=(40.08*g*mol^-1)/(100.09*g*mol^-1)=40%$ with respect...
1 Answers 1 viewsFirst, you want to find the ratio between the empirical formula and molecular formula; that is, how much has the molecular formula been simplified to reach the empirical formula. We...
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 viewsWe need (i) a stoichiometrically balanced equation.... $2LiOH(s) + CO_2(g)rarrLi_2CO_3(s) + H_2O$ And then (ii) we need equivalent quantities of carbon dioxide. There are.....$(1000*g)/(23.95*g*mol^-1)=417.5*mol$ with respect to lithium hydroxide....
1 Answers 1 views