How many moles of Calcium Nitrate would react with 0.7 moles of Chromium (III) Sulfate?;
The answer is 0.7 moles of Calcium Nitrate would react with 0.7 moles of Chromium (III) Sulfate.
To determine the number of moles of Calcium Nitrate that would react with 0.7 moles of Chromium (III) Sulfate, we need to look at the balanced chemical equation for the reaction between these two compounds.
The balanced chemical equation for the reaction between Calcium Nitrate (Ca(NO3)2) and Chromium (III) Sulfate (Cr2(SO4)3) is:
3 Ca(NO3)2 + 2 Cr2(SO4)3 → 6 CaSO4 + 2 Cr(NO3)3
From the balanced equation, we can see that 3 moles of Calcium Nitrate will react with 2 moles of Chromium (III) Sulfate.
Using the ratio of moles from the balanced equation, we can set up a proportion to find the number of moles of Calcium Nitrate:
(3 moles Ca(NO3)2 / 2 moles Cr2(SO4)3) = (x moles Ca(NO3)2 / 0.7 moles Cr2(SO4)3)
Cross-multiplying and solving for x, we get:
3 moles Ca(NO3)2 = (2 moles Cr2(SO4)3 * x moles Ca(NO3)2) / 0.7 moles Cr2(SO4)3
3 moles Ca(NO3)2 = 2.857 moles Ca(NO3)2
So, approximately 2.857 moles of Calcium Nitrate would react with 0.7 moles of Chromium (III) Sulfate.
To determine the number of moles of Calcium Nitrate (Ca(NO3)2) that would react with 0.7 moles of Chromium (III) Sulfate (Cr2(SO4)3), we need to compare the balanced chemical equation for their reaction.
The balanced chemical equation for the reaction between Calcium Nitrate and Chromium (III) Sulfate is:
3 Ca(NO3)2 + 2 Cr2(SO4)3 -> 6 CaSO4 + 2 Cr(NO3)3
By comparing the coefficients, we can see that the ratio of moles between Calcium Nitrate and Chromium (III) Sulfate is 3:2.
So, for every 3 moles of Calcium Nitrate, we need 2 moles of Chromium (III) Sulfate.
Now, if we have 0.7 moles of Chromium (III) Sulfate, we can calculate the number of moles of Calcium Nitrate required:
(0.7 moles of Cr2(SO4)3) * (3 moles of Ca(NO3)2 / 2 moles of Cr2(SO4)3) = 1.05 moles of Ca(NO3)2
Therefore, 0.7 moles of Chromium (III) Sulfate would react with 1.05 moles of Calcium Nitrate.