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3 years ago

PLEASE HELP! I DONT KNOW WHAT TO DO! ITS ALOT PLEASE HELP THO!

Chemistry

1 answer:

kirza4 [7]3 years ago

7 0

Answer:

CuCl2 (aq) + (NH4)2CO3(aq) → NH4Cl(aq) + CuCO3(s)

* CuCl2 (aq) + (NH4)2CO3(aq) → 2NH4Cl(aq) + CuCO3(s)

*Cu(2+)(aq)+ 2Cl(-)(aq) + 2NH4(+)(aq) + CO3(2-)(aq) →2NH4(+)(aq) + Cl(-)(aq) + CuCO3(s)

*Cu(2+)(aq)+ CO3(2-)(aq) → CuCO3(s)

*2HCl(aq) + 2MgCO3(s) → 2MgCl(aq) + H2O(l) + 2CO2(g)

* 2H(+)(aq) + 2Cl(-)(aq) + 2MgCO3(s) → 2Mg(2+)(aq) +2Cl(-) (aq) + H2O(l) + 2CO2(g)

* 2MgCO3(s) → 2Mg(2+)(aq) + H2O(l) + 2CO2(g)

*ZnCl2(aq) + 2AgC2H3O2(aq) → 2AgCl(s) + Zn(C2H3O2)2(aq)

*Zn(2+)(aq) + 2Cl(-)(aq) + 2Ag(+)(aq) + C2H3O2(-)(aq) → 2AgCl(s) + Zn(2+)(aq) 2C2H3O2(-)(aq)

*2Cl(-)(aq) + 2Ag(+)(aq) → 2AgCl(s)

*MnO(s) + H2SO4(aq) → MnSO4(s) + H2O (l)

*MnO(s) + 2H(+)(aq) + SO4(2-)(aq) → MnSO4(s) + H2O (l)

*FeS(s) + 2HCl (aq) → FeCl2(aq) + H2S(g)

*FeS(s) + 2H(+)(aq) + 2Cl(-)(aq) → Fe(2+)(aq)+ 2Cl(-)(aq)) + H2S(g)

*FeS(s) + 2H(+)(aq) → Fe(2+)(aq)+ H2S(g)

Explanation:

In the first equation you only need to know the rules of solubility. All chlorides are solubles except Ag+ Pb+2 and Hg2(2+) and most carbonates are insolubles

<em>For ionic equations</em> Start with a balanced molecular equation. Break all soluble strong electrolytes (compounds with (aq) beside them) into their ions indicate the correct formula and charge of each ion . Indicate the correct number of each ion and write (aq) after each ion . Bring down all compounds with (s), (l), or (g) unchanged.

<em>For net ionic equations </em>Write the balanced molecular equation. Write the balanced complete ionic equation. Cross out the spectator ions that are present. Write the "leftovers" as the net ionic equation.

Before this, te equation must be balanced.

* CuCl2 (aq) + (NH4)2CO3(aq) → 2NH4Cl(aq) + CuCO3(s)

*Cu(2+)(aq)+ 2Cl(-)(aq) + 2NH4(+)(aq) + CO3(2-)(aq) →2NH4(+)(aq) + Cl(-)(aq) + CuCO3(s)

*Cu(2+)(aq)+ CO3(2-)(aq) → CuCO3(s)

*2HCl(aq) + 2MgCO3(s) → 2MgCl(aq) + H2O(l) + 2CO2(g)

* 2H(+)(aq) + 2Cl(-)(aq) + 2MgCO3(s) → 2Mg(2+)(aq) +2Cl(-) (aq) + H2O(l) + 2CO2(g)

* 2MgCO3(s) → 2Mg(2+)(aq) + H2O(l) + 2CO2(g)

*ZnCl2(aq) + 2AgC2H3O2(aq) → 2AgCl(s) + Zn(C2H3O2)2(aq)

*Zn(2+)(aq) + 2Cl(-)(aq) + 2Ag(+)(aq) + C2H3O2(-)(aq) → 2AgCl(s) + Zn(2+)(aq) 2C2H3O2(-)(aq)

*2Cl(-)(aq) + 2Ag(+)(aq) → 2AgCl(s)

*MnO(s) + H2SO4(aq) → MnSO4(s) + H2O (l)

*MnO(s) + 2H(+)(aq) + SO4(2-)(aq) → MnSO4(s) + H2O (l)

*FeS(s) + 2HCl (aq) → FeCl2(aq) + H2S(g)

*FeS(s) + 2H(+)(aq) + 2Cl(-)(aq) → Fe(2+)(aq)+ 2Cl(-)(aq)) + H2S(g)

*FeS(s) + 2H(+)(aq) → Fe(2+)(aq)+ H2S(g)

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How much CaCO3 would have to be decomposed to produce 247 g of CaO

vovangra [49]

441 g CaCO₃ would have to be decomposed to produce 247 g of CaO

<h3>Further explanation</h3>

Reaction

Decomposition of CaCO₃

CaCO₃ ⇒ CaO + CO₂

mass CaO = 247 g

mol of CaO(MW=56 g/mol) :

$\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{247}{56}\\\\mol=4.41$

From equation, mol ratio CaCO₃ : CaO = 1 : 1, so mol CaO :

$\tt \dfrac{1}{1}\times 4.41=4.41$

mass CaCO₃(MW=100 g/mol) :

$\tt mass=mol\times MW\\\\mass=4.41\times 100\\\\mass=441~g$

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