Question

How many grams of calcium metal would be produced if you started the reaction with 10 grams of calcium borate?
How much of the excess reactant would be left over at the end of the reaction?
$6Li+C{{a}_{3}}{{(B{{O}_{3}})}_{2}}\xrightarrow{{}}3Ca+2L{{i}_{3}}B{{O}_{3}}$

Answer 1

Hint:To solve this question, we first need to understand the concept of mole ratio.
A conversion factor which relates the number of moles of any two components of a chemical reaction is known as the mole ratio. The coefficients in a balanced chemical equation of a reaction are used to determine the numbers in a conversion factor.

Complete step-by-step answer:Here, we can see that one mole of calcium borate produces three moles of calcium metal.
Now, we know that different relative amounts of reactants and products can also be considered by using the same proportionality (mole ratio concept).
Also, we know that the molar mass of calcium borate is 237.852 g/mol. And, the atomic weight of calcium metal is 40.078 u.
So, we can say that 273.852 g of calcium borate produces $3\times 40.078\text{ g}=120.234\text{ g}$ of calcium metal.
Hence, 10 g of calcium borate will produce
\[
 =10\times \dfrac{120.234}{237.852}g \\
\cong 5.05g\text{ Ca} \\
\]
So, approximately 5.05 grams of calcium metal would be produced if we start the reaction with 10 grams of calcium borate.
Now, we can see that 6 moles of lithium react with 1 mole of calcium borate to form the desired products.
The atomic mass of Li is 6.941 u.
So, $6\times 6.941=41.646\text{ }g$of lithium reacts with 273.852 g of calcium borate to form the desired products.
Hence, 10 g of calcium borate would require \[\dfrac{41.464}{273.852}\times 10\text{ g}\cong 1.52\text{ g}\] of lithium to form the desired products. The rest of the amount of excess lithium would be left over at the end of the reaction.

Note:The yield of products in any reaction are governed by the amount of the limiting reactant present. The limiting reactant will be completely utilized at the end of a chemical reaction. It is also known as a limiting reagent or a limiting agent.
In the given question, the amount of lithium was not specified and a fixed amount of calcium borate was given, hence in the reaction calcium borate acts as a limiting reagent and lithium is assumed to be the excess reagent.