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3 years ago

5

How many moles of calcium nitrate would react with 4.55 moles of chromium(iii) sulfate to produce calcium sulfate and chromium(i

ii) nitrate

2 answers:

Rasek [7]3 years ago

6 0

The equation for the reaction between the calcium nitrate and chromium sulfate to form calcium sulfate and chromium(iii) nitrate is,
3Ca(NO3)2 + Cr2(SO4)3 --> 3CaSO4 + 2Cr(NO3)3
The balanced equation shows that for every 1 mole of chromium sulfate, there will be 3 moles of calcium nitrate. Therefore for this item, there will be 4.55 moles times 3 which is equal to 13.65.

defon3 years ago

3 0

Answer:

The answer is 13.65 moles of Ca(NO3)2

Explanation:

the balanced equation is as follows:

3Ca(NO3)2 + Cr2(SO4)3 = 3CaSO4 + 2Cr(NO3)2

According to the balanced equation, 3 moles of Ca(NO3)2 react with 1 mole of Cr2(SO4)3. Using a simple rule of three, we can calculate the number of moles of Ca(NO3)2 that would react with 4 moles of Cr2(SO4)3. As follows:

3 moles Ca(NO3)2------------------- 1 mol Cr2(SO4)3

X moles Ca(NO3)2------------------- 4.55 moles Cr2(SO4)3

Clearing X:

X moles Ca(NO3)2 = 3*4.55 = 13.65 moles

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A 50.0 mL solution of 0.141 M KOH is titrated with 0.282 M HCl . Calculate the pH of the solution after the addition of each of

Kobotan [32]

Answer:

pH =1 2.84

Explanation:

First we have to start with the <u>reaction</u> between HCl and KOH:

$HCl~+~KOH->~H_2O~+~KCl$

Now <u>for example, we can use a volume of 10 mL of HCl</u>. So, we can calculate the moles using the <u>molarity equation</u>:

$M=\frac{mol}{L}$

We know that $10mL=0.01L$ and we have the concentration of the HCl $0.282M$ , when we plug the values into the equation we got:

$0.282M=\frac{mol}{0.01L}$

$mol=0.282*0.01$

$mol=0.00282$

We can do the same for the KOH values ( $50mL=0.05L$ and $0.141M$ ).

$0.141M=\frac{mol}{0.05L}$

$mol=0.141*0.05$

$mol=0.00705$

So, we have so far <u>0.00282 mol of HCl</u> and <u>0.00705 mol of KOH</u>. If we check the reaction we have a <u>molar ratio 1:1</u>, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an <u>excess of KOH</u>. This excess can be calculated if we <u>substract</u> the amount of moles:

$0.00705-0.00282=0.00423mol~of~KOH$

Now, if we want to calculate the pH value we will need a <u>concentration</u>, in this case KOH is in excess, so we have to calculate the <u>concentration of KOH</u>. For this, we already have the moles of KOH that remains left, now we need the <u>total volume</u>:

$Total~volume=50mL+10mL=60mL$

$60mL=0.06L$

Now we can calculate the concentration:

$M=\frac{0.00423mol}{0.06L}$

$M=0.0705$

Now, we can <u>calculate the pOH</u> (to calculate the pH), so:

$pOH=-Log(0.0705)$

$pOH=1.15$

Now we can <u>calculate the pH value</u>:

$14=~pH~+~pOH$

$pH=14-1.15=12.84$

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