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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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The reaction between $H_2$ and $O_2$ produces 13.1 g of water. How many grams of $O_2$ reacted? $2H_2(g) + O_2(g) -> 2H_2O(g)$?
There are $(13.1*g)/(18.01*g*mol^-1)=0.727*mol$ of water product. $H_2(g) + 1/2O_2(g) rarr H_2O(l)$ Given the of the reaction, there were necessarily $0.727*mol$ dihydrogen reactant, and $(0.727*mol)/2$ dioxygen reactant. I speak of dihydrogen,...
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For the decomposition reaction $2Ag_2CO_3->4Ag+2CO_2+O_2$, how many moles of reactant undergo decomposition in order to produce 6.0 moles of $Ag$?
The reaction above correctly represents thermal decomposition of a metal carbonate. The carbonate gives metal oxide and carbon dioxide under STRONG heating. You are not going to get silver metal...
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Which products form when $HgO$ decomposes? $2HgO (s) ->?$
$2HgO (s) ->2Hg(l)+O_2(g)$ HgO(s) when heated strongly metallic Hg and gaseous Oxygen are formed.
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$Mg$ and $O_2$ react in a 2.1 molar ratio. 2 moles $M_g$ = 1 mole $O_2$. If a reaction used 32.5 g of $O_2$, how many g of $Mg$ reacted?
Step 1. Calculate the moles of $"O"_2$. $32.5 cancel("g O"_2) × ("1 mol O"2)/(32.00 cancel("g O"2)) = "1.016 mol O"_2$ Step 2. Use the molar ratio to calculate the...
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In the formula $H_2O_2 -> H_2O + O_2$, how many grams of $O_2$ are produced from the decomposition of 68 g of $H_2O_2$?
Your equation is not correctly balanced. The correct equation is as follows: $2"H"_2"O"_2 -> 2"H"_2"O" + "O"_2$ First, calculate the moles of $"H"_2"O"_2$ reacting. In order to do this, we...
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What is the percent yield of $O_2$ if 10.2 g of $O_2$ is produced from the decomposition of 17.0 g of $H_2O$?
The can be calculated by: $%Yield=("Actual Yield")/("Theoretical Yield")xx100%$ The actual yield of oxygen is given and it is $10.2g$. We will need to find the theoretical yield. The decomposition reaction...
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A common laboratory preparation for small for quantities of $O_2$ is to decompose $KClO_3$ by heating: $2KClO_3 (s) -> 2KCl (s) + 3O_2 (g)$. If the decomposition of 2.00 g of $KCIO_3$ gives 0.720 g of $O_2$, what is the percent yield for the reaction?
Start by examining the balanced chemical equation that describes this $color(darkgreen)(2)"KClO"_ (3(s)) stackrel(color(white)(acolor(red)(Delta)aaa))(->) 2"KCl"_ ((s)) + color(blue)(3)"O"_ (2(g))$ Notice that it takes $color(darkgreen)(2)$ moles of potassium chlorate...
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$2H_2O -> 2H_2 + O_2$ What is the percent yield of $O_2$ if 10.2 g of $O_2$ is produced from the decomposition of 17.0 g of $H_2O$?
You know by looking at the balanced chemical equation $2"H"_ 2"O"_ ((l)) -> 2"H"_ (2(g)) + "O"_ (2(g))$ that it takes $2$ moles of water to produce...
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A sample containing 4.80 g of $O_2$ gas has a volume of 15.0 L at constant pressure and temperature. What is the new volume, in liters, after 0.500 mole of $O_2$ gas is added to the initial sample of 4.80 g of $O_2$?
Initially, you have $4.80color(red)(cancel(color(black)("g O"_2))) × "1 mol O"_2/(32.00 color(red)(cancel(color(black)("g O"_2)))) = "0.1500 mol O"_2$ Thus, you have $"0.1500 mol O"_2/"15.0 L" = "0.0100 mol O"_2/"1 L"$ If you...
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