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how do i solve these equations using proust's law? we haven't gone over balancing chem equations in class and our teacher doesn't want us to use those but instead some kind of ratio? can someone explain

how do i solve these equations using proust's law? we haven't gone over balancing-example-1
User Rosangela
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1 Answer

23 votes
23 votes

Answer:

Step-by-step explanation:

Ok so Proust's Law, better known as the Law of Definite Proportions, states that the components of a compound always exist in a fixed ratio. This means that it does not matter what the source of the components are nor the coefficients in front of them. The ratio your teacher is referring to is most likely mass percent/ percent composition. This ratio is the amount of the component over the amount of the entire compound times 100%.

*I am not very familiar with this law so please do with my answers what you will :)*

7.)

First, you want to find the percent composition (aka mass composition) of Na in NaCl.

45.89 g Na
--------------------- x 100% = 39.3%
116.89 g NaCl

So, there must be 39.3% sodium in NaCl. You can find how much chlorine is in NaCl by subtracting that percent by 100 (to find the percent composition of chlorine) then multiplying it by the mass.

100% - 39.3% = 60.7%

60.7% / 100 = 0.0607

116.89 g NaCl x 0.0607 = 70.91 g Cl₂

You could also just subtract the mass of sodium from the mass of sodium chloride to find the mass of chlorine.

116.89 g NaCl - 45.89 g Na = 71 g Cl₂

8.)

10.57 g Mg + 6.96 g O₂ = 17.53 g MgO

6.96 g O₂
--------------------- x 100% = 39.7% O₂
17.53 g MgO

9.)

6.46 g Pb = 1 g O₂

68.54 g Pb = 28.76 g O₂

68.54 g Pb / 28.76 g O₂ = 2.83 g ≠ 6.46 g

No, the two samples are not the same because the proportion of lead to oxygen is not the same for both samples. In the first sample, there is 6.46 g lead for every oxygen. In the second sample, there is 2.38 g lead for every oxygen.

User Emily Kothe
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