Chemistry, 17.09.2019 05:00 marshalldayton1452

Balance me those equations :
a=_fe + _ h2so4 --> _fe2 (so4)3 + _h2
b= _3b2 br6 + _ hno 3 --> _b(no3)3 +_hbr
c=_bf3+_li2so3 > b2(so3)3 + lif
d=_(nh4) 3 po4 + _pb(no3)4 > pb3(po4)4+_nh4no3
e= _secl6 + _o2 --> _seo2+ _ cl2
you don't got to fill up every " _ ".
after balancing the equations from above , explain me , how? i do not get it .

Answers: 2

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Answer from: lays20001

In balancing equations, we aim to get equal numbers of every type of atom on both sides of the equation, in order to satisfy the law of conservation of mass (which states that in a chemical reaction, every atom in the reactants is reorganised to form products, without exception). Therefore, let me walk you through question a:

_Fe + _ H2SO4 --> _Fe2 (SO4)3 + _H2

First, take a stock-check of exactly what we currently have on each side (assuming that each _ represents a 1):

LHS: Fe = 1, H = 2, S = 1, O = 4
RHS: Fe = 2, H = 2, S = 3, O = 12,

There are two things to note here. Firstly, H2 (it should be subscript in reality) represents two hydrogen atoms bonded together as part of the ionic compound H2SO4 (sulphuric acid) - this two only applies to the symbol which is directly before it. Hence, H2SO4 only contains 1 sulphur atom, because the 2 applies to the hydrogen and the 4 applies to the oxygen. Secondly, the bracket before the 3 (which should also be subscript) means that there is 3 of everything within the bracket - (SO4)3 contains 3 sulphur atoms and 12 oxygen atoms (4 * 3 = 12).

Now let's start balancing. As a prerequisite, you must keep in mind that we can only add numbers in front of whole molecules, whereas it is not scientifically correct to change the little numbers (we could have two sulphuric acids instead of one, represented by 2H2SO4 (where the 2 would be a normal-sized 2 when written down), but we couldn't change H2SO4 to H3SO4).

The iron atoms can be balanced by having two iron atoms on the left-hand side instead of one:

2Fe + _ H2SO4 --> _Fe2 (SO4)3 + _H2

Now let's balance the sulphur atoms, by multiplying H2SO4 by 3:

2Fe + 3H2SO4 --> _Fe2 (SO4)3 + _H2

This has the added bonus of automatically balancing the oxygens too. This is because SO4- is an ion, which stays the same in a displacement reaction (which this one is). Take another stock check:

LHS: Fe = 2, H = 6, S = 3, O = 12
RHS: Fe = 2, H = 2, S = 3, O = 12

The only mismatch now is in the hydrogen atoms. This is simple to rectify because H2 appears on its own on the right-hand side. Just multiply H2 by 3 to finish off, and fill the third gap with a 1 because it has not been multiplied up. Alternatively, you can omit the 1 entirely:

2Fe + 3H2SO4 --> Fe2 (SO4)3 + 3H2

This is the balanced symbol equation for the displacement of hydrogen with iron in sulphuric acid.

For question b, I will just show you the stages without the explanation (I take the 3 before B2 to be a mistake, because it makes no sense to use 3B2Br6 when B2Br6 balances fine):

B2 Br6 + _ HNO 3 -->_B(NO3)3 +_HBr
B2Br6 + _HNO3 --> _B(NO3)3 + 6HBr
B2Br6 + 6HNO3 --> _B(NO3)3 + 6HBr
B2Br6 + 6HNO3 --> 2B(NO3)3 + 6HBr

Hopefully you can get the others now yourself. I hope this helped

Answer from: peacelillady1030

2Fe + 3H2SO4 --> Fe2 (SO4)3 + 3H2
Done the first but i have to go sorry

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

Answer from: Quest

Iimpact me by the one the yet but him.

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Balance me those equations :
a=_fe + _ h2so4 --> _fe2 (so4)3 + _h2
b= _3b2 br6 +...

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