# chemistry

calculate the amount of 95% marble required to obtain 11.2Litres OF CARBON DIOXIDE

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1. CaCO3 + 2HCl ==> CaCl2 + H2O + CO2
11.2 L CO2 (atSTP) = 11.2/22.4 = 0.5 mol.
1 mole CO2 in the reaction = 1 mole CaCO3 initially; therefore, 0.5 mole CO2 means you started with 0.5 mol CaCO3.
grams = moles x molar mass = ??
Since the marble is only 95% pure, you must start with a little more marble or
??g CaCO3 calculated/0.95 = xx.
Here is a worked example of a stoichiometry problem.
http://www.jiskha.com/science/chemistry/stoichiometry.html

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2. You should calculate fully and try to be a little more understanble

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3. CaCO3→CaO+CO2CaCO3→CaO+CO2

Given volume of CO2" role="presentation" style="box-sizing: border-box; display: inline-block; line-height: 0; text-indent: 0px; text-align: left; text-transform: none; font-style: normal; font-weight: normal; font-size: 19.2px; letter-spacing: normal; overflow-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; CO2CO2 = 11.2 litres

Molar volume at STP = 22.4 litres

1&#xA0;mole&#xA0;CO2&#xA0;in&#xA0;the&#xA0;reaction&#xA0;=&#xA0;1&#xA0;mole&#xA0;CaCO3No.&#xA0;of&#xA0;moles&#xA0;of&#xA0;CO2&#xA0;=&#xA0;frac11.222.4=0.5moles" role="presentation" style="box-sizing: border-box; display: table-cell !important; line-height: 0; text-indent: 0px; text-align: center; text-transform: none; font-style: normal; font-weight: normal; font-size: 19.36px; letter-spacing: normal; overflow-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 20.935em; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; width: 10000em; position: relative;">1 mole CO2 in the reaction = 1 mole CaCO3No. of moles of CO2 = frac11.222.4=0.5moles1 mole CO2 in the reaction = 1 mole CaCO3No. of moles of CO2 = frac11.222.4=0.5moles

1 mole &#xA0;rmCO2" role="presentation" style="box-sizing: border-box; display: inline-block; line-height: 0; text-indent: 0px; text-align: left; text-transform: none; font-style: normal; font-weight: normal; font-size: 19.2px; letter-spacing: normal; overflow-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; position: relative;"> rmCO2 rmCO2 in the reaction = 1 mole CaCO_3

Therefore, 0.5 mole CO2 means 0.5 mol CaCO3 has reacted in the above reaction.

No.&#xA0;of&#xA0;grams&#xA0;of&#xA0;CaCO3=no.&#xA0;of&#xA0;moles&#xA0;of&#xA0;CaCO3&#x00D7;molar&#xA0;mass&#xA0;of&#xA0;CaCO3=0.5&#x00D7;100=50g" role="presentation" style="box-sizing: border-box; display: table-cell !important; line-height: 0; text-indent: 0px; text-align: center; text-transform: none; font-style: normal; font-weight: normal; font-size: 19.36px; letter-spacing: normal; overflow-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 31.886em; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; width: 10000em; position: relative;">No. of grams of CaCO3=no. of moles of CaCO3×molar mass of CaCO3=0.5×100=50gNo. of grams of CaCO3=no. of moles of CaCO3×molar mass of CaCO3=0.5×100=50g

Since the marble is only 95% pure,

Amount of marbleCaCO3=500.95=52.63g" role="presentation" style="box-sizing: border-box; display: inline-block; line-height: 0; text-indent: 0px; text-align: left; text-transform: none; font-style: normal; font-weight: normal; font-size: 19.2px; letter-spacing: normal; overflow-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; position: relative;">CaCO3=500.95=52.63g

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