1. Ca(OH)2 + H3PO4 ---> H2O + Ca3(PO4)2
(A) How many moles of H3PO4 are needed to make 15g of H2O?
(B) What mass in grams of Ca(OH)2 is used to react with 5.6mol of H3PO4?
(C) How many grams of Ca3(PO4)2 are made from 134g of Ca(OH)2?
2. Fe2O3 + CO ---> Fe + CO2
(A) If 110g Fe2O3 is allowed to react with 80g CO - which is the limiting reactant?
(B) How much Fe can be made? - answer in grams.
(C) How much excess reactant is left over?
3. NO + O2 --> NO2
(A) If 5 mol of NO and 2 mol of O2 are combined who limits the reaction?
(B) How many moles of the excess reactant remain unused?
(C) How much moles NO2 can be produced?
Optional 4. C + S8 --> CS2
(A) If I mix 16g of C and 70g of S8, what is the limiting reactant?
(B) How much product can be made? - answer in moles
(C) How much excess reactant will remain after the experiment? - answer in grams
http://www.jiskha.com/science/chemistry/stoichiometry.html
http://www.chem.tamu.edu/class/majors/tutorialnotefiles/limiting.htm
(A) To find the number of moles of H3PO4 needed to make 15g of H2O, we first need to determine the molar mass of H2O. H2O has a molar mass of 18 g/mol.
Next, we need to determine the stoichiometry of the reaction between Ca(OH)2 and H3PO4. According to the balanced equation, the molar ratio between H3PO4 and H2O is 1:3.
Now, we can calculate the number of moles of H3PO4 needed:
15g H2O * (1 mol H2O/18g H2O) * (1 mol H3PO4/3 mol H2O) = 0.278 moles of H3PO4
Therefore, approximately 0.278 moles of H3PO4 are needed to make 15g of H2O.
(B) To find the mass in grams of Ca(OH)2 used to react with 5.6 moles of H3PO4, we first need to determine the molar mass of Ca(OH)2. Ca(OH)2 has a molar mass of 74 g/mol.
Next, we need to determine the stoichiometry of the reaction between Ca(OH)2 and H3PO4. According to the balanced equation, the molar ratio between Ca(OH)2 and H3PO4 is 1:2.
Now, we can calculate the mass of Ca(OH)2 used:
5.6 moles H3PO4 * (1 mol Ca(OH)2/2 mol H3PO4) * (74 g Ca(OH)2/1 mol Ca(OH)2) = 206.8 grams of Ca(OH)2
Therefore, approximately 206.8 grams of Ca(OH)2 are used to react with 5.6 moles of H3PO4.
(C) To find the number of grams of Ca3(PO4)2 made from 134g of Ca(OH)2, we first need to determine the molar mass of Ca3(PO4)2. Ca3(PO4)2 has a molar mass of 310 g/mol.
Next, we need to determine the stoichiometry of the reaction between Ca(OH)2 and Ca3(PO4)2. According to the balanced equation, the molar ratio between Ca(OH)2 and Ca3(PO4)2 is 1:1.
Now, we can calculate the mass of Ca3(PO4)2 made:
134g Ca(OH)2 * (1 mol Ca3(PO4)2/1 mol Ca(OH)2) * (310 g Ca3(PO4)2/1 mol Ca3(PO4)2) = 4194 grams of Ca3(PO4)2
Therefore, approximately 4194 grams of Ca3(PO4)2 are made from 134g of Ca(OH)2.
To solve these questions, we need to use stoichiometry, which involves calculating the amount of reactants needed and the amount of products produced based on the balanced chemical equation.
1. Ca(OH)2 + H3PO4 ---> H2O + Ca3(PO4)2
(A) To calculate the moles of H3PO4 needed to make 15g of H2O, we first need to find the molar mass of H2O, which is 18.015 g/mol. Next, we can set up a mole-to-mole ratio using the balanced equation. From the equation, we can see that the ratio of H3PO4 to H2O is 1:3. Therefore, the moles of H3PO4 needed can be calculated as follows:
moles of H2O = mass of H2O / molar mass of H2O
moles of H3PO4 = 3 * moles of H2O
(B) To find the mass in grams of Ca(OH)2 used to react with 5.6 mol of H3PO4, we need to use the mole-to-mole ratio between Ca(OH)2 and H3PO4 from the balanced equation, which is 1:1. Therefore, the mass of Ca(OH)2 can be calculated as follows:
mass of Ca(OH)2 = moles of H3PO4 * molar mass of Ca(OH)2
(C) To determine the grams of Ca3(PO4)2 produced from 134g of Ca(OH)2, we need to use the mole-to-mole ratio between Ca(OH)2 and Ca3(PO4)2 from the balanced equation, which is 1:1. Therefore, the grams of Ca3(PO4)2 can be calculated as follows:
grams of Ca3(PO4)2 = 134g of Ca(OH)2 * (molar mass of Ca3(PO4)2 / molar mass of Ca(OH)2)
2. Fe2O3 + CO --> Fe + CO2
(A) To determine the limiting reactant between 110g of Fe2O3 and 80g of CO, we need to calculate the moles of each reactant. Use the molar mass of each compound to calculate the moles, and then compare the moles to the stoichiometric ratio in the balanced equation. The reactant that produces fewer moles of product is the limiting reactant.
moles of Fe2O3 = mass of Fe2O3 / molar mass of Fe2O3
moles of CO = mass of CO / molar mass of CO
(B) To find the mass of Fe produced, we need to use the mole-to-mole ratio between Fe2O3 and Fe from the balanced equation, which is 2:3. Therefore, the mass of Fe can be calculated as follows:
moles of Fe = 3/2 * moles of Fe2O3
mass of Fe = moles of Fe * molar mass of Fe
(C) To find the excess reactant remaining, we compare the moles of the excess reactant to the stoichiometric ratio in the balanced equation. Subtract the moles of the limiting reactant used from the total moles of the excess reactant to determine the remaining moles.
3. NO + O2 --> NO2
(A) To determine which reactant limits the reaction between 5 mol of NO and 2 mol of O2, we need to compare the moles of each reactant to the stoichiometric ratio in the balanced equation. The reactant that does not have enough moles to react completely is the limiting reactant.
(B) To calculate the moles of the excess reactant remaining, we subtract the moles of the limiting reactant used from the total moles of the excess reactant.
(C) To find the moles of NO2 produced, use the mole-to-mole ratio between NO and NO2 from the balanced equation, which is 2:2. Therefore, the moles of NO2 can be calculated as follows:
moles of NO2 = moles of NO
Optional 4. C + S8 --> CS2
(A) To determine the limiting reactant between 16g of C and 70g of S8, we need to calculate the moles of each reactant and compare them to the stoichiometric ratio in the balanced equation. The reactant that produces fewer moles of product is the limiting reactant.
moles of C = mass of C / molar mass of C
moles of S8 = mass of S8 / molar mass of S8
(B) To find the moles of CS2 produced, use the mole-to-mole ratio between C and CS2 from the balanced equation, which is 1:1. Therefore, the moles of CS2 can be calculated as follows:
moles of CS2 = moles of C
(C) To find the excess reactant remaining, we compare the moles of the excess reactant to the stoichiometric ratio in the balanced equation. Multiply the moles of the excess reactant remaining by its molar mass to determine the grams remaining.