How to find the theoretical yield for two solid reactants?

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How to find the theoretical yield for two solid reactants?

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Answer 1 · 2018-04-26T22:34:34

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Explanation:

Lets say you want to calculate the theoretical yield for the reaction of 10 grams of hydrogen gas reacting with oxygen gas forming water.

First start by balancing the chemical equation:
$H_{2} (g) + O_{2} (g) \to H_{2} O (l)$ $H_2(g)+O_2(g) -> H_2O(l)$

This gives
$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l)$ $2H_2(g)+O_2(g) -> 2H_2O(l)$

Now we determine the mole ratio of the reactants and the product.
For this reaction two mole hydrogen gas gives two mole water, therefore the mole ratio between $H_{2} (g)$ $H_2(g)$ and $H_{2} O (l)$ $H_2O(l)$ is one. $(\frac{1 m o l H_{2}}{1 m o l H_{2} O})$ $((1 m o l H_2)/(1 m o l H_2O))$

Now we can calculate the theoretical yield using the molar mass of the reactant to convert grams of reactant to moles of reactant.

Molar mass:

$H_{2} = 2 g$ $H_2 = 2 g$
$H_{2} O = 18 g$ $H_2O = 18 g$

Grams of water:

$m_{H_{2} O} = 10 g H_{2} \cdot (\frac{1 m o l H_{2}}{2 g H_{2}} \cdot (\frac{1 m o l H_{2} O}{1 m o l H_{2}}) \cdot (\frac{18 g H_{2} O}{1 m o l H_{2} O}))$ $m_(H_2O)=10 g H_2* ((1 mol H_2)/(2 g H_2)* ((1 mol H_2O)/(1 mol H_2)) * ((18 g H_2O)/(1 m o l H_2O)))$

this gives:

$m_{H_{2} O} = 90 g$ $m_(H_2O) = 90 g$

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How to find the theoretical yield for two solid reactants?

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