The mystery salt imagine that you have a barrel of salt, but you forgot to label it. you know it must be either kno3, or kcl. you look at the solubility curves for kno3 and kcl and you find that at 35 degrees celsius, 100 g of water can dissolve about 30 g of kno3, or about 37 g of kcl. you have a scale, a hot plate, a thermometer, empty beakers, and plenty of water. 1. how can you determine whether the barrel contains kno3 or kcl? hint: what property could you use to identify the mystery salt? (1 point) 2. explain exactly what you would do and why you would do it. (2 points) 3. how would your results identify the mystery salt? hint: what results would indicate kno3? what results would indicate kcl? (2 points)

Question 1 :

To determine whether the salt is KCl or KNO3, one should look for the difference between them in terms of their physicochemical properties, such as their solubility.

Since we have the solubilité of KCl and KNO3, we can use the property of solubility to determine if the mystery salt is KCl or KNO3.

Question 2:

We will try to reproduce the conditions to determine the solubility of the salt at 37°C.

We will put into the beaker 100ml of water (equivalent to 100g) and dissolve a defined quantity of the salt (the number should be between the solubility of the KCl (37g) and KNO3 (30g) so between 30g and 37g).

Let's dissolve for example 32g of the salt, then, heat with the hotplate until the temperature of the beaker content will be 35 ° C (use the thermometer to determine the exact temperature).

Why?

This manipulation aims to determine the solubility of our mystery salt to know if it is KNO3 or KCl. In our conditions, we will obtain two different possibilities depending on if the salt is KCl or KNO3, this justifies why we took a quantity between 30g and 37g of salt.

If it is KNO3 (solubility of 30g/ml) we will observe a precipitation in the beaker because we exceed its solubility.

If it is KCl (37g/100) we will not observe a precipitate since we did not attempt the solubility of KCl

Question 3:

Finally to determine the composition of salt: we know that the solubility of KCL is 37g / 100ml (that is to say if we dissolve a higher mass (38g for example), we will observe a precipitation of salt) and that the solubility of KNO3 is 30g / 100ml (that is to say if we dissolve an upper mass (32g for example), we will observe a precipitation of salt)

In our experiment, 32g of salts were dissolved. If it is KCl, we will not observe a precipitate since the minimum concentration to start having a precipitate is not yet reached (37g / 100ml).

If it is KNO3, a precipitate will be observed since the minimum concentration to start having a precipitate is not yet reached (30g / 100ml).

is there any options? maybe im just not seeing them but i need the options to

-love shrooms

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2