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Elden [556K]
3 years ago
8

A compound of a transition metal and iodine is 56.7% metal by mass.How many grams of the metal can be obtained from 630 g of thi

s compound?
Chemistry
1 answer:
erma4kov [3.2K]3 years ago
4 0

Answer:

357 g of the transition metal are present in 630 grams of the compound of the transition metal and iodine

Explanation:

In any sample of the compound, the percentage by mass of the transition metal is 56.7%. This means that for a 100 g sample of the compound, 56.7 g is the metal while the remaining mass, 43.3 g is iodine.

Given mass of sample compound = 630 g

Calculating the mass of iodine present involves multiplying the percentage by mass composition of the metal by the mass of the given sample;

56.7 % = 56.7/100 = 0.567

Mass of transition metal = 0.567 * 630 = 357.21 g

Therefore, the mass of the transition metal  present in 630 g of the compound is approximately 357 g

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3 years ago
15.50 g of NH4Cl reacts with an excess of AgNO3. In the reaction 35.50 g AgCl is produced. what is the theoretical yield of AgCl
Readme [11.4K]

Answer:

The answer to your question is 41.6 g of AgCl

Explanation:

Data

mass of NH₄Cl = 15.5 g

mass of AgNO₃ = excess

mass of AgCl = 35.5 g

theoretical yield = ?

Process

1.- Write the balanced chemical reaction.

              NH₄Cl  +  AgNO₃   ⇒   AgCl  +  NH₄NO₃

2.- Calculate the molar mass of NH₄Cl and AgCl

NH₄Cl = 14 + 4 + 35.5 = 53.5 g

AgCl = 108 + 35.5 = 143.5 g

3.- Calculate the theoretical yield

                 53.5 g of NH₄Cl -------------------- 143.5 g of AgCl

                  15.5 g of NH₄Cl  -------------------    x

                         x = (15.5 x 143.5) / 53.5

                         x = 2224.25 / 53.5

                         x = 41.6 g of AgCl

8 0
3 years ago
Aqueous hydrobromic acid hbr will react with solid sodium hydroxide naoh to produce aqueous sodium bromide nabr and liquid water
Natalka [10]
 Balanced equation is

HBr + NaOH ----> NaBr + H2O

Using molar masses

80.912 g HBr reacts with  39.997 g of Naoh to give 18.007 g water

so 1 gram of NaOH reacts with 2.023 g of HBR   
and 5.7 reacts with 11.531 g HBr so we have excess HBr in this reaction

Mass  of water produced  =    (5.7 * 18.007 / 39.997  =  2.6 g to 2 sig figs
8 0
3 years ago
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The pOH of a solution is 6.0. Which statement is correct? Use p O H equals negative logarithm StartBracket upper O upper H super
castortr0y [4]

Answer:

The pH of the solution is 8.0.

Explanation:

taking the test rn

6 0
3 years ago
Read 2 more answers
When 6.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of H2 can be formed? 2Al + 6HCl → 2AlCl
4vir4ik [10]

Well, we need to find the ratio of Al to the other reactant.


Al:HCl = 1:3


--> this means that for every 1 Al used, you have to use 3 HCl.



6*3 = 18 moles of HCl needed to fully react with 6 moles of Al. Since 13<18, HCL is the limiting reactant.



The ratio of HCl:AlCl = 3:1



13/3 = 4.3333...



The final answer is HCl is the limiting reactant with 4.3 moles of AlCl3 able to be produced.



Hope this helps!!! :)


6 0
3 years ago
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