Question

$Ca{Cl}_2+{Be}_2{CO}_3\to Ca{CO}_3+Be{Cl}_3$
$I$. Provided the reaction above, if $1.10g$ of $Ca{Cl}_2$ and $0.85g$ of ${Be}_2{CO}_3$ will be the limiting reactant.
$II$. The limiting reactant in a reaction will always be the reactant that has the lowest mass.

• A. Statement $I$ is true, Statement $II$ is true
• B. Statement $I$ is true, Statement $II$ is false
• C. Statement $I$ is false, Statement $II$ is true
• D. Statement $I$ is false, Statement $II$ is false

Answer 1

The correct option is D Statement $I$ is false, Statement $II$ is false

Molar mass of $CaC{l}_2=110g$

Molar mass of $B{e}_{2}C{O}_3=78g$

$CaC{l}_2+B{e}_{2}C{O}_3\to CaC{O}_3+BeC{l}_2+2H_{2}O$

$1.10g0.85g$

Here, no. of moles of $B{e}_{2}C{O}_3$ are high compared to $CaC{l}_2.$ So, $CaC{l}_2$ is consumed completely.

Limiting reagent is reactant which gets consumed completely. So, $CaC{l}_2$ is limiting reagent.