Question-227418

Question Number 227418 by Spillover last updated on 24/Jan/26

Answered by Kassista last updated on 25/Jan/26

a) At constant temperature and pressure, equal volumes of all gases contain the same number of molecules. ⇒ (V/n) = constant b) A mole is the amount of substance that contais 6.022∙10^(23) elementary entities, such as atoms and ions c) CaCO_(3 (s)) +2HNO_(3 (l)) → CO_(2 (g)) + Ca(NO_3 )_(2 (aq)) + H_2 O_((l)) ∴ 2 mol of HNO_3 − 1 mol CaCO_3 M_(CaCO_3 ) =100(g/(mol)) ⇒ 2 mol HNO_3 − 1 mol CaCO_3 x mol HNO_3 − 0.05 mol CaCO_3 ⇒ x=0.1 mol HNO_3

$$\left.{a}\right)\:{At}\:{constant}\:{temperature}\:{and}\:{pressure},\:{equal}\:{volumes} \\ $$$${of}\:{all}\:{gases}\:{contain}\:{the}\:{same}\:{number}\:{of}\:{molecules}. \\ $$$$\Rightarrow\:\frac{{V}}{{n}}\:=\:{constant} \\ $$$$\left.{b}\right)\:{A}\:{mole}\:{is}\:{the}\:{amount}\:{of}\:{substance}\:{that}\:{contais} \\ $$$$\mathrm{6}.\mathrm{022}\centerdot\mathrm{10}^{\mathrm{23}} \:{elementary}\:{entities},\:{such}\:{as}\:{atoms}\:{and}\:{ions} \\ $$$$ \\ $$$$\left.{c}\right)\:{CaCO}_{\mathrm{3}\:\left({s}\right)} +\mathrm{2}{HNO}_{\mathrm{3}\:\left({l}\right)} \:\rightarrow\:{CO}_{\mathrm{2}\:\left({g}\right)} \:+\:{Ca}\left({NO}_{\mathrm{3}} \right)_{\mathrm{2}\:\left({aq}\right)} \:+\:{H}_{\mathrm{2}} {O}_{\left({l}\right)} \\ $$$$\therefore\:\mathrm{2}\:{mol}\:{of}\:{HNO}_{\mathrm{3}} \:−\:\mathrm{1}\:{mol}\:{CaCO}_{\mathrm{3}} \\ $$$$\:\:\:\:\:{M}_{{CaCO}_{\mathrm{3}} } =\mathrm{100}\frac{{g}}{{mol}} \\ $$$$\Rightarrow\:\mathrm{2}\:{mol}\:{HNO}_{\mathrm{3}} \:−\:\mathrm{1}\:{mol}\:{CaCO}_{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:{x}\:{mol}\:{HNO}_{\mathrm{3}} \:−\:\mathrm{0}.\mathrm{05}\:{mol}\:{CaCO}_{\mathrm{3}} \\ $$$$\Rightarrow\:{x}=\mathrm{0}.\mathrm{1}\:{mol}\:{HNO}_{\mathrm{3}} \\ $$

Commented by Spillover last updated on 01/Feb/26

$${wrong} \\ $$

Commented by Kassista last updated on 02/Feb/26

The question asked the amount of the substance nitric acid, that′s meadured with moles, the mass would be 6.3g (indeed 0.1 mol)

$${The}\:{question}\:{asked}\:{the}\:{amount}\:{of}\:{the}\:{substance} \\ $$$${nitric}\:{acid},\:{that}'{s}\:{meadured}\:{with}\:{moles}, \\ $$$${the}\:{mass}\:{would}\:{be}\:\mathrm{6}.\mathrm{3}{g}\:\left({indeed}\:\mathrm{0}.\mathrm{1}\:{mol}\right) \\ $$

Answered by Spillover last updated on 01/Feb/26

CaCO_(3 (s)) +2HNO_(3 (l)) → CO_(2 (g)) + Ca(NO_3 )_(2 (aq)) + H_2 O_((l)) From balanced chemical equation Molar mass CaCO_3 =100g/mol Molar mass 2HNO_3 =126g/mol 100g→126g 5g→x x=((5×126)/(100))=6.30g HNO_3 6.30g HNO_3

$$\:{CaCO}_{\mathrm{3}\:\left({s}\right)} +\mathrm{2}{HNO}_{\mathrm{3}\:\left({l}\right)} \:\rightarrow\:{CO}_{\mathrm{2}\:\left({g}\right)} \:+\:{Ca}\left({NO}_{\mathrm{3}} \right)_{\mathrm{2}\:\left({aq}\right)} \:+\:{H}_{\mathrm{2}} {O}_{\left({l}\right)} \\ $$$${From}\:{balanced}\:{chemical}\:{equation} \\ $$$${Molar}\:{mass}\:{CaCO}_{\mathrm{3}} =\mathrm{100}{g}/{mol} \\ $$$${Molar}\:{mass}\:\mathrm{2}{HNO}_{\mathrm{3}} =\mathrm{126}{g}/{mol} \\ $$$$\mathrm{100}{g}\rightarrow\mathrm{126}{g} \\ $$$$\mathrm{5}{g}\rightarrow{x} \\ $$$${x}=\frac{\mathrm{5}×\mathrm{126}}{\mathrm{100}}=\mathrm{6}.\mathrm{30}{g}\:{HNO}_{\mathrm{3}} \\ $$$$\mathrm{6}.\mathrm{30}{g}\:{HNO}_{\mathrm{3}} \\ $$

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