Question

What mass of hydrogen is produced?
$100$ g of lithium metal is dropped into $1$ L of water. What mass of hydrogen is produced?
a) $14\cdot 3$ g
b) $28\cdot 6$ g
c) $15\cdot 4$ g
d) $18\cdot 6$ g

Answer 1

Hint : We are going to use the concept of stoichiometry and limiting reagent and calculating the solution.
Limiting reagent is the reactant that is used up completely in the reaction, and stops the product formation.
Stoichiometric coefficient is the constant number written in front of atoms/molecules.

Complete Step By Step Answer:
Writing the correct the balanced equation: - $2$ Li + $2$ H $_{2}$ O $\to$ $2$ LiOH + H $_{2}$
Determine the amount of Li and water in moles i.e.
Moles of Li= mass in gram /molecular weight in gram $\Rightarrow$ $100/6\cdot 941$ = $14\cdot 40$ mol.
Moles of H $_{2}$ O = mass in gram/ molecular weight in gram $\Rightarrow$ $1000/18=55.55$ mol.
Determine the limiting reagent of the reactant (for that we need to divide the number of moles of the reactant by their stoichiometric coefficient) i.e., moles of Li= $14\cdot 40/2=7\cdot 20$ mol and H $_{2}$ O= $55\cdot 55/2=27\cdot 77$ mol.
From above calculation we came to know that Li is the limiting reagent and product formation depends upon moles if Li.
Mass of $7\cdot 20$ mole of dihydrogen gas = mole * molecular weight of H $_{2}$ .
Mass = $7\cdot 20$ mol * $2$ gram per mol = $14.4$ gram.
So, the correct answer is $14\cdot 4$ gram, Option (A).

Note :
You have to carefully calculate the solution step by step, by calculating the number of moles of the reactant and divide it by their stoichiometric coefficients to find the limiting reagent. Once you find the limiting reagent you should start finding the product formation as per asked in the question.