calculate the amount of co2 produced when 100 grams of calcium carbonate is treated with 36.5 grams of HCL
Answers
Generated mass of \bold{\mathrm{CO}_{2}}CO
2
during in the given reaction is 8.8 g.
Explanation:
Balanced reaction: \mathrm{CaCO}_{3}+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_{2}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}CaCO
3
+2HCl→CaCl
2
+CO
2
+H
2
O
From the reaction,
1 mole of calcium carbonate reacts with 2 moles of Hydrogen chloride and produce 1 mole of carbon dioxide.
Let’s calculate the moles of \mathrm{CaCO}_{3}CaCO
3
¬:
Molar mass of \mathrm{CaCO}_{3}CaCO
3
= 100 g/mol
Given mass of \mathrm{CaCO}_{3}CaCO
3
= 20 g
Moles of \mathrm{CaCO}_{3}=\frac{\text { Mass of } \mathrm{CaCo}_{3}}{\text { Molar mass of } \mathrm{CaCo}_{3}}=\frac{20}{100}=0.20\ \mathrm{mol}CaCO
3
=
Molar mass of CaCo
3
Mass of CaCo
3
=
100
20
=0.20 mol
0.20 mol of \bold{\mathrm{CaCO}_{3}}CaCO
3
reacts with 0.4 mol of HCl
Let’s calculate the moles of HCl:
Molar mass of HCl = 36.45 g/mol
Given mass = 20 ml
Moles of HCl= \frac{\text { Mass of }\mathrm{HCl}}{\text {Molar mass of }\mathrm{HCl}}=\frac{20}{36.45}=0.548\ \mathrm{mol}
Molar mass of HCl
Mass of HCl
=
36.45
20
=0.548 mol
Therefore, \mathrm{CaCO}_{3}CaCO
3
is limiting reagent.
0.20 mol of \mathrm{CaCO}_{3}CaCO
3
gives 0.20 mol of \mathrm{CO}_{2}CO
2
Molar mass of \mathrm{CO}_{2}CO
2
= 44 g/mol
Mass of \mathrm{CO}_{2}\ produced = 0.2 \times 44 \frac{g}{m o l}=8.8 gCO
2
produced=0.2×44
mol
g
=8.8g
Hope it helps u
Mark me as brainliest
꧁༺DҽѵíӀ ցíɾӀ༻꧂