QUESTION 1:
In this reaction:
CaCO3 + 2HCl ---> CaCl2 + CO2 + H2O
6.088 g CaCO3 reacted with 2.852 g HCl. What mass of CaCO3 remains unreacted?
SOLUTION :
HCl is Limiting reagent
CaCO3 ⇒ 6.088 g / 100.086 g/mol = 0.0608277 mol
HCl ⇒ 2.852 g / 36.461 g/mol = 0.0782206 mol
Grams of calcium carbonate used:
1 is to 2 as x is to 0.0782206 mol
x = 0.0391103 mol
0.0391103 mol times 100.086 g/mol = 3.914 g
Grams of CaCO3 remaining: 6.088 g minus 3.914 g = 2.174 g
QUESTION 2:
950.0 grams of copper(II) sulfate are reacted with 460.0 grams of zinc metal. (a) What is the theoretical yield of Cu? (b) If 295.8 grams of copper are actually obtained from this reaction, what is the percent yield?
SOLUTION:
The balanced chemical equation: CuSO4 + Zn ---> ZnSO4 + Cu
According to the balanced chemical equation, calculate the moles of reactants.
CuSO4 ⇒ 950.0 g / 159.607 g/mol = 5.95212 mol
Zn ⇒ 460.0 g / 65.38 g/mol = 7.03579 mol
So, CuSO4 is limiting reagent.
Grams of Cu: 5.95212 mol of Cu is produced (as 1 : 1 molar ratio involved)
5.95212 mol X 63.546 g/mol = 378.2 g
Percent yield : 295.8 g / 378.2 g = 78.21 %
QUESTION 3:
Silicon nitride (Si3N4) is made by combining Si and nitrogen gas (N2) at a high temperature. How much (in g) Si is needed to react with an excess of nitrogen gas to prepare 125 g of silicon nitride if the percent yield of the reaction is 95.0%?
Answer: 79.1 g Si
QUESTION 4:
Use the following reaction: C4H9OH + NaBr + H2SO4 C4H9Br + NaHSO4 + H2O
If 15.0 g of C4H9OH reacts with 22.4 g of NaBr and 32.7 g of H2SO4 to yield 17.1 g of C4H9Br, what is the percent yield of this reaction?
Answer: 61.5 %.
QUESTION 5:
If 4.95 g of ethylene (C2H4) are combusted with 3.25 g of oxygen.
a. What is the limiting reagent?
b. How many grams of CO2 are formed?
Answer: Oxygen and 2.98 g CO2
QUESTION 6:
If you dissolve lead(II) nitrate and potassium iodide in water they will react to form lead(II) iodide and potassium nitrate.
a.Write the equation for this reaction, then balance the equation.
b.Calculate the grams of lead(II) iodide that can be produced from 5.00 moles of potassium iodide.
c.Calculate the grams of lead(II) iodide that can be produced from 75.00 grams of potassium iodide.
Answer: 1.15 x 103 g PbI2 and 104.1 g PbI2
QUESTION 7:
Write then balance the combustion reaction for propane gas, C3H8.
a. If 5.00 grams of propane burn completely, what volume of carbon dioxide is produced at STP?
b. If 75.0 L of steam is produced at STP, what mass of propane must have burned?
c. If 34.2 grams of propane are completely combusted, how many moles of steam will that produce?
Answer: 7.62 L CO2 , 36.9 g C3H8 and 3.10 mol steam
QUESTION 8:
If 31.6 grams of magnesium hydroxide is combined with 68.0 mL of 0.725 M HBr, which is the limiting reagent? How many grams of magnesium bromide would be formed?
Answer: HBr is the limiting reagent and 4.54 g MgBr2
QUESTION 9:
Suppose 316.0 g aluminum sulfide reacts with 493.0 g of water. What mass of the excess reactant remains?
The unbalanced equation is:
Al2S3 + H2O ---> Al(OH)3 + H2S
Answer: Al2S3 is the limiting reagent.227.4685 g of water and 265.5 g excess.
QUESTION 10:
What weight of each substance is present after 0.4500 g of P4O10 and 1.5000 g of PCl5 are reacted completely?
P4O10 + 6PCl5 ---> 10POCl3
Answer: PCl5 is limiting, 0.1092 g of P4O10 remaining, 1.8408 g of POCl3 produced
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