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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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In the reaction $2H_2O -> 2H_2 + O_2$, if you start with 2 mol of water, how many moles of hydrogen gas are produced?
This is a fairly simple example of a titration. We use titrations to calculate how many (grams, moles, litres, molecules, etc.) of one compound/element we would have if we are...
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What is the number of moles of $O_2$ that will react with 10.0g of $H_2$ to form water in the equation $2H_2 + O_2 -> 2H_2O$?
For this problem, we should implement the process of . First, convert the grams of $H_2$ into moles: $10.0g H_2xx(1molH_2)/(2.01588 g H_2)$ Then multiply by your mole to mole ratio:...
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Given the equation $2H_2O ->2H_2 + O_2$, how many moles of $H_2O$ would be required to produce 2.5 moles of $O_2$?
Since the chemical equation, $color(red)2H_2O->2H_2+color(blue)1O_2$ requires a specific number of moles of the reactant, then a specific number of moles of the products are created. With this knowledge,...
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Given the equation $2H_2O -> 2H_2 + O_2$, how many moles of $H_2O$ be required to produce 2.5 moles of $O_2$?
So every 2 molecules of water gives 1 molecule of oxygen. This ratio ($2:1$) stays the same if we are talking about moles, because a mole is a set number...
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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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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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For the reaction: $2H_2O + 137 kcal -> 2H_2(g) + O_2(g)$ how many kcal are needed to form 2.00 moles $O_2(g)$?
Start by taking a look at the thermochemical equation given to you $2"H"_ 2"O"_ ((l)) + "137 kcal" -> 2"H"_ (2(g)) + "O"_ (2(g))$ Notice that this thermochemical...
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