1. Ca(OH)2(aq) + H3PO4(aq) à Ca3(PO4)2(aq) + H2O(l)
a. If you have 13.7mol of Ca(OH)2, how many grams of H2O are produced?
b. How many grams of H3PO4 are required to produce 102.3g Ca3(PO4)2?
c. How many moles of Ca(OH)2 are needed to produce 82.9g of Ca3(PO4)2?
2. Al2(SO4)3 + Ca(OH)2à Al(OH)3 + CaSO4
a. What mass of Ca(OH)2 is needed to react with 132.8g Al2(SO4)3?
b. How many molecules of CaSO4 are produced if 145g of Al(OH)3 are also produced?
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To solve these problems, we need to use stoichiometry, which involves using balanced chemical equations and molar ratios to determine the quantities of substances involved in a chemical reaction.
1. a. To find the grams of H2O produced, we need to first determine the balanced equation and the molar ratio between Ca(OH)2 and H2O.
The balanced equation is:
Ca(OH)2(aq) + H3PO4(aq) -> Ca3(PO4)2(aq) + H2O(l)
From the balanced equation, we can see that for every 1 mole of Ca(OH)2, 1 mole of H2O is produced. Therefore, the molar ratio is 1:1.
Given that we have 13.7 mol of Ca(OH)2, we know that 13.7 mol of H2O will be produced.
To find the mass of H2O produced, we need to use the molar mass of H2O, which is 18.015 g/mol.
Mass of H2O = 13.7 mol * 18.015 g/mol = 246.6 g
Therefore, 246.6 grams of H2O are produced.
b. To find the grams of H3PO4 required, we need to again use the balanced equation and the molar ratio between Ca3(PO4)2 and H3PO4.
The molar ratio between Ca3(PO4)2 and H3PO4 is 1:2, meaning that for every 1 mole of Ca3(PO4)2, 2 moles of H3PO4 are required.
Given that we have 102.3 g of Ca3(PO4)2, we can set up a proportion to find the number of moles of H3PO4 required.
102.3 g Ca3(PO4)2 / (molar mass of Ca3(PO4)2) = x moles H3PO4 / (molar mass of H3PO4)
Therefore, the grams of H3PO4 required can be calculated as:
grams of H3PO4 = (102.3 g * molar mass of H3PO4) / (molar mass of Ca3(PO4)2)
c. To find the moles of Ca(OH)2 needed, we again use the balanced equation and the molar ratio between Ca(OH)2 and Ca3(PO4)2.
The molar ratio between Ca(OH)2 and Ca3(PO4)2 is 1:1, meaning that for every 1 mole of Ca(OH)2, 1 mole of Ca3(PO4)2 is produced.
Given that we have 82.9 g of Ca3(PO4)2, we can set up a proportion to find the number of moles of Ca(OH)2 needed.
82.9 g Ca3(PO4)2 / (molar mass of Ca3(PO4)2) = x moles Ca(OH)2 / (molar mass of Ca(OH)2)
Therefore, the moles of Ca(OH)2 needed can be calculated as:
moles of Ca(OH)2 = (82.9 g * molar mass of Ca(OH)2) / (molar mass of Ca3(PO4)2)
2. a. To find the mass of Ca(OH)2 needed, we again use the balanced equation and the molar ratio between Al2(SO4)3 and Ca(OH)2.
The molar ratio between Al2(SO4)3 and Ca(OH)2 is 1:3, meaning that for every 1 mole of Al2(SO4)3, 3 moles of Ca(OH)2 are required.
Given that we have 132.8 g of Al2(SO4)3, we can set up a proportion to find the number of moles of Ca(OH)2 needed.
132.8 g Al2(SO4)3 / (molar mass of Al2(SO4)3) = x moles Ca(OH)2 / (molar mass of Ca(OH)2)
Therefore, the grams of Ca(OH)2 needed can be calculated as:
grams of Ca(OH)2 = (132.8 g * molar mass of Ca(OH)2) / (molar mass of Al2(SO4)3)
b. To find the number of molecules of CaSO4 produced, we use the molar mass and Avogadro's number.
First, calculate the moles of Al(OH)3 produced using the given mass and molar mass.
Then, use the balanced equation to find the molar ratio between Al(OH)3 and CaSO4. The ratio is 1:1, meaning that for every 1 mole of Al(OH)3 produced, 1 mole of CaSO4 is produced.
Finally, multiply the number of moles of CaSO4 by Avogadro's number to find the number of molecules.
This explanation should help you solve the given problems by using stoichiometry and the principles of balanced chemical equations.