8. calculate the volume of co₂ at stp formed from the decomposition of 152 grams of caco₃. 34.05l

Explanation:

Step1: Write the decomposition equation

$CaCO_3(s)\xrightarrow{\Delta}CaO(s)+CO_2(g)$

Step2: Calculate the molar mass of $CaCO_3$

The molar mass of $Ca$ is 40.08 g/mol, $C$ is 12.01 g/mol and $O$ is 16.00 g/mol. So, $M_{CaCO_3}=40.08 + 12.01+3\times16.00=100.09$ g/mol.

Step3: Calculate the number of moles of $CaCO_3$

$n_{CaCO_3}=\frac{m}{M}=\frac{152\ g}{100.09\ g/mol}\approx1.52$ mol

Step4: Determine the mole - ratio

From the balanced equation, the mole - ratio of $CaCO_3$ to $CO_2$ is 1:1. So, $n_{CO_2}=n_{CaCO_3}=1.52$ mol

Step5: Use the molar volume at STP

At STP (Standard Temperature and Pressure, $T = 273\ K$ and $P= 1\ atm$), the molar volume of any ideal gas is $V_m = 22.4$ L/mol. $V_{CO_2}=n_{CO_2}\times V_m=1.52\ mol\times22.4\ L/mol = 34.048\ L\approx34.05$ L

Answer:

34.05 L