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And given the experiment, we would find half an equiv of dihydrogen gas evolved per equiv of metal, and one equiv of hydroxide in solution. Both results could be easily confirmed by experiment.
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Which of the given equations correctly represents the neutralization of magnesium hydroxide by hydrochloric acid?
$"a. "Mg(OH)_2(s)+2HCl(aq) rarr MgCl_2(aq) + 2H_2O(l)$
$"b. "Mg(OH)_2(s)+2HCl(aq) rarr MgO(s) + 2H_2(g) + Cl_2(g)$
$"c. "MgO(s)+2HCl(aq)$
We need a stoichiometric equation that represents the reaction between $Mg(OH)_2$ and $HCl$, and here it is............ $Mg(OH)_2(s) + 2HCl(aq) rarr MgCl_2(aq) + 2H_2O(l)$ Here, $1$ $"equiv"$ of magnesium hydroxide...
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Is $2H_2(g) + O_2(g) rarr 2H_2O(l)$ the same as $H_2(g) + 1/2O_2(g) rarr H_2O(l)$?
$2H_2(g)+O_2(g) rarr 2H_2O(l)$ $36*g$ of reactants, and $36*g$ of products.......... VERSUS....... $H_2(g) + 1/2O_2(g)rarr H_2O(l)$ $18*g$ of reactants, and $18*g$ of products.......... Is mass conserved in each reaction? Well, clearly...
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Which equation is correctly balanced:
$A.$ $Al(s) + 3O_2(g) rarr Al_2O_3(s);$
$B.$ $Al(s) + 3O_2(g) rarr Al_2O_3(s);$
$C.$ $Al(s) + 3/2O_2(g) rarr Al_2O_3(s);$
$D.$ $2Al(s) + 3O_2(g) rarr 2Al_2O_3(s)?$
Alumina is $Al_2O_3$, which is the product of aluminum oxidation. None of your equations shows the correct formulae. We can rewrite the correct equation as: $2Al(s)+ 3/2O_2(g) rarr Al_2O_3(s)$ This...
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This is the equation for burning hydrogen in air: $2H_2 + O_2 -> 2H_2O. How many H and O atoms are shown on the left-hand side of the equation?
Clearly, there are 4 hydrogen atoms, but 2 hydrogen molecules, and 2 oxygen atoms, but 1 oxygen molecule on the left hand side of the equation. And on the RHS,...
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How many moles of tin metal will be produced when 0.74 moles of tin IV oxide is heated in the reaction $SnO_2 + 2H_2 -> Sn + 2H_2O$?
So, why clearly? Well, to begin, you already have the stoichiometric equation: $SnO_2(s) +2H_2(g) rarr Sn(s) + 2H_2O(l)$ This is a so-called $"redox reaction"$: stannic oxide is reduced to tin...
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Water can be formed from the stoichiometric reaction of hydrogen with oxygen: $2H_2(g) + O_2(g) + 2H_2O(g)$. A complete reaction of 5.0 g of $O_2$ with excess hydrogen produces how many grams of $H_2O$?
$2H_2(g) + O_2(g) rarr 2H_2O(g)$ We started with $(5.0*g)/(32.00*g*mol^-1)=0.156*mol" dioxygen"$. Given excess dihydrogen, this molar quantity thus gives $2xx0.156*mol$ water. And thus, $2xx0.156*molxx18.01*g*mol^-1~=6*g$ water are evolved. Is this reaction exothermic...
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Which equation is unbalanced?
$HCl(aq)+H_2O(l)rarr Cu^(2+) + 2NO_3^-$;
$Ca(OH)_2(s) stackrel(H_2O)rarr Ca^(2+) +2HO^-$;
$Ca(NO_3)_2(s) stackrel(H_2O)rarr Ca^(2+) +2NO_3^-$
;$NH_3(aq) +H_2O(l) rightleftharpoonsNH_4^+ + HO^-$;
$HCl(aq)+H_2O(l)rarr Cu^(2+) + 2NO_3^-$ $Ca(OH)_2(s) stackrel(H_2O)rarr Ca^(2+) +2HO^-$ $Ca(NO_3)_2(s) stackrel(H_2O)rarr Ca^(2+) +2NO_3^-$ $NH_3(aq) +H_2O(l) rightleftharpoonsNH_4^+ + HO^-$ And one of these things is not like the other ones.........
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How does ammonia behave as a Bronsted base:
$"(i) "NH_3(g) + 7/2O_2(g) rarr 2NO_2(g) + 3H_2O$
$"(ii) "2NH_3(g) rarr NH_4^+ +NH_2^(-) $
$"(iii) "NH_3(g) + H_3CBr(g) rarr H_3CBr(g) + HBr(g)$
$"(iv) "NH_3(aq) +H_2O(l) rarr NH_4^+ + HO^-$?
Ammonia is a Bronsted base: $NH_3(aq) + H_2O(l) rightleftharpoons NH_4^+ + OH^-$ $pK_b = 4.75$ I am not going to solve this equation using $pK_b$, but most of the ammonia...
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For which equation, does the enthalpy of reaction represent $DeltaH_f^@$ for methanol?
$1.$ $CO(g) + 2H_2(g) rarr CH_3OH(l)$ $2.$ $C(g) + 2H_2O(g) rarr CH_3OH(l)+1/2H_2(g)$ $3.$ $C(s) + 2H_2(g) +1/2O_2(g) rarr CH_3OH(l)$ $4.$ $C(g) + 2H_2(g) rarr CH_3OH$
The standard of formation is the enthalpy associated with the formation of 1 mole of substance from its constituent in their standard states under specified conditions. And of course...
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When 2 moles of $H_2(g)$ and 1 mole of $O_2(g)$ react to give liquid water, 572 kJ of heat evolve.
$2H_2(g)$ + $O_2(g)$ $rarr$ $2H_2O(l)$ ; $DeltaH$ = -572 kJ
a) What is the themochemical eq. for 1 mole of liquid water?
You know that when $2$ moles of hydrogen gas react with $1$ mole of oxygen gas, you get $2$ moles of water and $"572 kJ"$ of heat are evolved....
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