Chemistry - Stoichiometry Question with Solution | TestHub

ChemistryStoichiometryBasic Methods of Calculations (POAC, LR)Medium2 minQB

ChemistryMediummultiple choice

50 gm of $CaCO_{3}$ is allowed to react with 68.6 gm of $H_{3} PO_{4}$ then select the correct option( s $) : (Ca = 40, P = 31) CaCO_{3} + H_{3} PO_{4} \to Ca_{3} (PO_{4})_{2} + H_{2} O + CO_{2}$

Options:(select one or more)

Answer:

A, B

Solution:

$3 C a C O_{3} + 2 H_{3} P O_{4} \to C a_{3} (P O_{4})_{2} + 3 H_{2} O + 3 C O_{2}$

W : 50g 68.6

n : 0.5 0.7

SA : $\frac{0.5}{3}$ $\frac{0.7}{2}$

Therefore, $C a C O_{3}$ is the limiting reagent.

Stream:JEESubject:ChemistryTopic:StoichiometrySubtopic:Basic Methods of Calculations (POAC, LR)

⏱ 2mℹ️ Source: QB

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