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Is $2H_2(g) + O_2(g) rarr 2H_2O(l)$ the same as $H_2(g) + 1/2O_2(g) rarr H_2O(l)$?

Md Ariful Islam

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Md Ariful Islam

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$2H_2(g)+O_2(g) rarr 2H_2O(l)$

$36*g$ 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 mass IS conserved. And while of course I cannot have half a molecule of dioxygen gas, I can certainly have a $16*g$ quantity of $O_2(g)$, i.e. a HALF A MOLE quantity of dioxygen gas. The use of such fractional coefficients often makes the arithmetic involved in a bit easier, which of course is why they are used.

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When 1.57 mol $O_2$ reacts with $H_2$ to form $H_2O$, how many moles of $H_2$ are consumed in the process?

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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 $H_2O$ 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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