Ninjakat Posted September 23, 2010 Posted September 23, 2010 Im unsure what im supposed to do. Please dont just be blunt im actually lost here.. I need direction. 25. The following reaction represents the decompostion of water: 2 H2O ---> 2H2 + O2 How many molecules of hydrogen are produced from the decompostion of 12.2 g of water into its elements? (4 marks) 26. Determine the mass of carbon monoxide that is produced when 45.6 g of methane, CH4, react with 73.2g of oxygen gas, O2. The products are carbon monoxide, CO, and water vapour H2O. (7 marks) 1
dttom Posted September 23, 2010 Posted September 23, 2010 (edited) So you may begin with calculating number of mole of substances produced per reaction, and calculate how many reactions have been performed. Edited September 23, 2010 by dttom
Ninjakat Posted September 23, 2010 Author Posted September 23, 2010 So you may begin with calculating number of mole of substances produced per reaction, and calculate how many reactions have been performed. What question are you talking about? And what am I using? like.. i said no blunt comments be specific. 1
ewmon Posted September 23, 2010 Posted September 23, 2010 25. The reaction: 2·H2O —► 2·H2 + O2 using 12.2 g of water a. What is the definition of a “mole”? b. How many moles of water are in the problem? c. What's the ratio of hydrogen gas molecules produced per water molecule? d. How many moles of hydrogen gas are produced? (This thought sequence will help to prepare you for #26)
wanabe Posted September 23, 2010 Posted September 23, 2010 Have you attempted a "unit line equation" aka "dimensional analysis"? let's see your setup.
Ninjakat Posted September 23, 2010 Author Posted September 23, 2010 Hey guys, someome on the chat portion of the website was able to help me last night , but thanks for your time 25. h20 = 17.99 g/mol nh2o = m/M = 12.2g / 17.99 g/mol =0.67 moles Since 1 H2O = 1 H2 = 1:1 ratio H = 0.67 moles 26. 2 CH + 3 O2 --> 2CO+4H2O nCH4 = 45.6g / 16.01 g/mol = 2.85 nO2 = 73.2g/31.98 g/mol = 2.28 3:2 Ratio, Oxygen is the limited reactant. Since CO is equal to 28 g/mol 2.28 / 3 moles of oxygen x 2 moles = 1.52 moles 1.52 x 28 = 42.56 moles. 2
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now