Balance the following equations Recall the stoichiometric coefficient (the numbe

Question

Balance the following equations Recall the stoichiometric coefficient (the numbers before the individual species)in a balanced chemical reaction refer to the number of motes of that species involved m me chemical reaction A balanced chemical equation shows the molar relationship between all of the reactants and products Now. given the Mowing chemical equation Is this equation balanced? (Yes/No) How many moles of HNO, are required to react with 6 moles of Cu? How many molos of Cu(NO*)a are produced when 3 moles of Cu react? How many motes of Cu must react to produce 4 motes of NO? If 4 motes of NO are produced, how many motes of H_2O are also produced? Propane. commonly used as a fuel in gas grills It reacts with oxygen according to How many grams of propanol

Explanation / Answer

1)

2 H2 + 1 02 ----> 2H20

2)

2 Mg + 1 02 ---> 2 MgO

3)

1 SF4 + 2 H20 ----> 1 S02 + 4 HF

4)

1 P4 + 5 02 ----> 2 P205

5)

1 C3H8 + 5 02 ---> 3 CO2 + 4 H2O

6)

a) yes it is balanced

b)

we can see that

moles of HN03 required = (8/3) x moles of Cu

so

moles of HN03 required = 8 x 6 / 3 = 16

c)

moles of Cu(N03)2 produced = moles of Cu reacted = 3

d)

moles of NO produced = (2/3) x moles of Cu

4 = (2/3) x moles of Cu

moles of Cu = 6

e)

moles of H20 produced = 2 x moles of NO produced = 2 x 4 = 8

7)

we know that

moles = mass / molar mass

so

moles of O2 = 24 / 32 = 0.75

the equation is

C3H8 + 502 ---> 3 CO2 + 4H2O

we can see that

moles of Propane burned = (1/5) x moles of O2

so

moles of propane burned = 0.75 / 5 = 0.15

now

mass = moles x molar mass

so

mass = 0.15 x 44 = 6.6

so

6.6 grams of propane can be burned

8)

moles of NH3 = 0.105 / 17 = 6.17647 x 10-3

we can see that

moles of Pb formed = 1.5 x moles of NH3 reacted

moles of Pb formed = 1.5 x 6.17647 x 10-3 = 9.2647 x 10-3

now

mass = moles x molar mass

so

mass of Pb formed = 9.2647 x 10-3 x 207.2 = 1.92

so

1.92 grams of Pb is formed

9)

the reaction is

2 LiOH + CO2 ----> Li2C03 + H20

we can see that

moles of LiOH required = 2 x moles of C02

so

moles of LiOH required = 2 x 0.08 = 0.16

so

0.16 moles of LioH is required

but

only 0.15 moles of LiOH is present

so

LiOH is the limiting reagent

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