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Which reactant will be used up first if 78.1g of o2 is reacted with 62.4g of c4h10?

dlinn [17]

Answer:

Reagent O₂ will be consumed first.

Explanation:

The balanced reaction between O₂ and C₄H₁₀ is:

2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

Then, by reaction stoichiometry, the following amounts of reactants and products participate in the reaction:

C₄H₁₀: 2 moles
O₂: 13 moles
CO₂: 8 moles
H₂O: 10 moles

Being:

C: 12 g/mole
H: 1 g/mole
O: 16 g/mole

The molar mass of the compounds that participate in the reaction is:

C₄H₁₀: 4*12 g/mole + 10*1 g/mole= 58 g/mole
O₂: 2*16 g/mole= 32 g/mole
CO₂: 12 g/mole + 2*16 g/mole= 44 g/mole
H₂O: 2*1 g/mole + 16 g/mole= 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of reactants and products participate in the reaction:

C₄H₁₀: 2 moles* 58 g/mole= 116 g
O₂: 13 moles* 32 g/mole= 416 g
CO₂: 8 moles* 44 g/mole= 352 g
H₂O: 10 moles* 18 g/mole= 180 g

If 78.1 g of O₂ react, it is possible to apply the following rule of three: if by stoichiometry 416 g of O₂ react with 116 g of C₄H₁₀, 62.4 g of C₄H₁₀ with how much mass of O₂ do they react?

$mass of O_{2} =\frac{416grams of O_{2}*62.4 grams ofC_{4}H_{10} }{116 grams of C_{4}H_{10}}$

mass of O₂= 223.78 grams

But 21.78 grams of O₂ are not available, 78.1 grams are available. Since you have less mass than you need to react with 62.4 g of C₄H₁₀, <u><em>reagent O₂ will be consumed first.</em></u>

3 0

3 years ago

Complete and balance the chemical equations for the precipitation reactions, if any, between the following pairs of reactants, a

Crank

Explanation:

a. $Pb(NO_3)_2(aq) + Na_2SO_4(aq)$ → ?

$Pb(NO_3)_2(aq) + Na_2SO_4(aq)\rightarrow PbSO_4(s)+2NaNO_3(aq)$

$Pb(NO_3)_2(aq)\rightarrow Pb^{2+}(aq)+2NO_3^{-}(aq)$

$Na_2SO_4(aq)\rightarrow 2Na^++SO_4^{2-}(aq)$

$Pb^{2+}(aq)+2NO_3^{-}(aq)+2Na^++SO_4^{2-}(aq)\rightarrow PbSO_4(s)+2Na^++2NO_3^{-}(aq)$

Removing common ions from both sides, we get the net ionic equation:

$Pb^{2+}(aq)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)$

b. $NiCl_2(aq) + NH_4NO_3(aq)$ →

$NiCl_2(aq) + NH4NO_3(aq) \rightarrow Ni(NO_3)_2+NH_4Cl(aq)$

No precipitation is occuring.

c. $Fe_Cl2(aq) + Na_2S(aq)$ →

$FeCl_2(aq) + Na_2S(aq)\rightarrow FeS(s)+2NaCl(aq)$

$FeCl_2(aq)\rightarrow Fe^{2+}(aq)+2Cl^{-}(aq)$

$Na_2S(aq)\rightarrow 2Na^++S{2-}(aq)$

$Fe^{2+}(aq)+2Cl^{-}(aq)+2Na^++S^{2-}(aq)\rightarrow FeS(s)+2Na^++2Cl^{-}(aq)$

Removing common ions from both sides, we get the net ionic equation:

$Fe^{2+}(aq)+S^{2-}(aq)\rightarrow FeS(s)$

d. $MgSO_4(aq) + BaCl_2(aq)$ →

$MgSO4(aq) + BaCl2(aq)\rightarrow BaSO_4(s)+MgCl_2$

$MgSO_4(aq)\rightarrow Mg^{2+}(aq)+SO_4^{2-}(aq)$

$BaCl_2(aq)\rightarrow Ba^{2+}+2Cl^{-}(aq)$

$Mg^{2+}(aq)+SO_4^{2-}(aq)+Ba^{2+}+2Cl^{-}(aq)(aq)\rightarrow BaSO_4(s)+Mg^{2+}(aq)+2Cl^{-}(aq)$

Removing common ions from both sides, we get the net ionic equation:

$Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

5 0

3 years ago

In a common experiment in the general chemistry laboratory, magnesium metal is heated in air to produce MgO. MgO is a white soli

Digiron [165]

Explanation:

When magnesium metal burns is heated i the air it forms magnesium oxide.The balanced chemical reaction is given as:

$2Mg+O_2\rightarrow 2MgO$

2 moles of magnesium metal when reacts with 1 moles of oxygen it gives 2 moles of magnesium oxide which is white in color.

Some times along with formation of magnesium oxide small amount of magnesium nitride also produced due to which magnesium oxide appears grey in color .The balanced chemical reaction is given as:

$3Mg+N_2\rightarrow Mg_3N_2$

3 moles of magnesium combines with 1 mol of nitrogen gas to to give 1 mol of magnesium nitride.

6 0

3 years ago

A bag of potato chips contains 585 mL of air at 20.0 C and a pressure of 765 mmHg. Assuming the bag does not break, what will be

Scorpion4ik [409]

Answer:

volume = 972.23ml

Explanation:

using general gas law

P1V1/T1 = P2V2/T2

765 x 585/293 = 443 x V2/282

1527.39 =443 x V2/282

1527.38 x 282 = 443 x V2

430695.78 = 443 x V2

V2 = 430695.68/443

V2 = 972.23mL

4 0

3 years ago

If a gas at 25oC occupies 3.60 L at a pressure of 2.50 atm, what will be its volume at a pressure of 1.00 atm if the temperature

Elenna [48]

Answer:

9L

Explanation:

Given parameters:

Initial volume V₁ = 3.6L

Initial pressure P₁ = 2.5atm

Final pressure P₂ = 1atm

Unknown:

Final volume V₂ = ?

Condition: constant temperature = 25°C

Solution:

This problem compares the volume and pressure of a gas at constant temperature.

This is highly synonymous to the postulate of Boyle's law. It states that "the volume of a fixed mass of gas is inversely proportional to the pressure provided that temperature is constant".

Mathematically;

P₁V₁ = P₂V₂

where P and V are pressure and volume

1 and 2 are initial and final states

Input the parameters and solve for V₂;

2.5 x 3.6 = 1 x V₂

V₂ = 9L

4 0

3 years ago