Thank you for being helpful in the past. Could you check my interpretation of the three chemical word problems below,and,let me know if I'm on the right track?
1.Determine the mass of water vapor formed when 1.00g of Butane (C4H10)is burned in a lighter(combined with Oxygen)to produce water and Carbon Dioxide.
C4H10(aq) + O2(g) ---> 5 H2O(g) + 4 CO2(g)
(I HAVE BALANCED PROBLEM ONE AS WELL)
2.How much Barium Chloride would be necessary to react with Silver Nitrate,if you have 5.0g of Silver Nitrate ?
BaCl2(aq) + AgNO3(aq) ----> Ba(NO3)2 (s) + AgCl(aq)
3. What mass of precipitate should form if 2.00g of Silver Nitrate is reacted with excess sodium sulfide solution?
2AgNO3(aq) + Na2S(aq) ---> Ag2S(aq) + NaNO3(s)
THANK YOU.
I don't think the equations are balanced (although, in some cases, the part that needs to be balanced is ok). However, it is good practice to have ALL of the equation balanced.
I have balanced them below and in bold face type I have correct the phases.
2C4H10(gas) + 13O2(g) ---> 10 H2O(g) + 8CO2(g)
BaCl2(aq) + 2AgNO3(aq) ----> Ba(NO3)2 (aq) + 2AgCl(s)
2AgNO3(aq) + Na2S(aq) ---> Ag2S(s) + 2NaNO3(aq)
Sure! Let's go through each problem one by one to check your interpretations.
1. Mass of water vapor formed:
To determine the mass of water vapor formed when 1.00g of butane is burned, we need to use the balanced equation: C4H10(aq) + O2(g) -> 5 H2O(g) + 4 CO2(g)
The coefficient in front of H2O tells us that for every mole of C4H10 burned, five moles of water vapor are formed. To determine the moles of C4H10, we divide the given mass (1.00g) by the molar mass of C4H10 (58.12 g/mol). This gives us the number of moles of C4H10.
Since the molar ratio between C4H10 and H2O is 1:5, we can multiply the number of moles of C4H10 by 5 to determine the number of moles of H2O. Then, multiply the moles of H2O by its molar mass (18.02 g/mol) to find the mass of water vapor formed.
2. Amount of Barium Chloride required:
To determine the amount of barium chloride required to react with 5.0g of silver nitrate, we need to use the balanced equation: BaCl2(aq) + AgNO3(aq) -> Ba(NO3)2 (s) + AgCl(aq)
The coefficient in front of AgNO3 tells us that for every mole of AgNO3, we need one mole of BaCl2. First, determine the moles of AgNO3 by dividing the given mass (5.0g) by its molar mass (169.87 g/mol). This gives us the number of moles of AgNO3.
Since the molar ratio between AgNO3 and BaCl2 is 1:1, the number of moles of BaCl2 required will also be equal to the moles of AgNO3. Multiply the moles of BaCl2 by its molar mass (208.23 g/mol) to find the mass of BaCl2 required.
3. Mass of precipitate formed:
To determine the mass of precipitate formed when 2.00g of silver nitrate reacts with excess sodium sulfide solution, we need to use the balanced equation: 2AgNO3(aq) + Na2S(aq) -> Ag2S(aq) + 2NaNO3(s)
The coefficient in front of Ag2S tells us that for every two moles of AgNO3, one mole of Ag2S is formed. First, determine the moles of AgNO3 by dividing the given mass (2.00g) by its molar mass (169.87 g/mol). This gives us the number of moles of AgNO3.
Since the molar ratio between AgNO3 and Ag2S is 2:1, divide the moles of AgNO3 by 2 to determine the moles of Ag2S formed. Then, multiply the moles of Ag2S by its molar mass (247.8 g/mol) to find the mass of precipitate formed.
I hope this clarifies your interpretations. Let me know if you have any further questions!