Respuesta :
Answer: 2800 g
Explanation:
[tex]CaCO_3(s)\rightarrow CaO(s)+CO_2(g)[/tex]
According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass = 5 kg = 5000 g
[tex]\text{Number of moles}=\frac{5000g}{100g/mol}=50moles[/tex]
1 mole of [tex]CaCO_3[/tex] produces = 1 mole of [tex]CaO[/tex]
50 moles of [tex]CaCO_3[/tex] produces =[tex]\frac{1}{1}\times 50=50moles[/tex] of [tex]CaO[/tex]
Mass of [tex]CaO=moles\times {\text{Molar mass}}=50moles\times 56g/mole=2800g[/tex]
2800 g of [tex]CaO[/tex] is produced from 5.0 kg of limestone.
A decomposition reaction splits the reactants into two or more products. The mass of the quicklime produced from 5 kg of limestone is 2800 gm.
What is mass?
Mass is the amount or the weight of the substance occupied in the system. It can be calculated in grams or kilograms.
The decomposition reaction of the Limestone can be shown as:
[tex]\rm CaCO_{3} \rightarrow CaO + CO_{2}[/tex]
The number of the mole of limestone is given as:
[tex]\rm Moles = \rm \dfrac {Mass}{Molar \;mass}[/tex]
Here, mass is 5000 gm and the molar mass is 100 g/mol
Substituting values in the equation above:
[tex]\begin{aligned}\rm n &= \dfrac{5000}{100}\\\\&= 50\;\rm mol\end{aligned}[/tex]
The stoichiometry coefficient of the reaction gives:
1 mole of limestone = 1 mole quicklime
So, 50 moles of limestone = x moles of quicklime
Solving for x:
[tex]\begin{aligned}\rm x &= \dfrac{50 \times 1}{1}\\\\&= 50 \;\rm mole\end{aligned}[/tex]
Mass of quicklime is calculated as:
[tex]\begin{aligned}\rm mass &= \rm moles \times \rm molar \; mass\\\\&= 50 \times 56\\\\&= 2800\;\rm gm\end{aligned}[/tex]
Therefore, 2800 gm of quicklime is produced.
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