Explanation:
a) HNO2(aq) = HNO3(aq) + H2O(l) +NO(g)
b) SoCl2 (l) + H2O (l) = So2(g) + 2HCl(aq)
c) CH4 (g) + 2O2(g) = Co2 (g) + 2H2O(g)
d) 3CuO(s) + 2NH3 (g) = 3Cu(s) + 3H2O (l) + N2(g)
Answer:
1 - 3
Explanation:
- Look to see where metals are on the periodic table then look at what group they are in. The group tells you the number of valence electrons. Ex. 1A has one valence electron.
- Hope this helped! If you need a further explanation please let me know.
Aluminum atomic number 13 becomes Neon with an atomic number of 10 when it loses three valence electrons.
Omg i lost everything ugh
To do it again
1. 12g+2(16g)= 44g/mol
25.01/ 44g/mol= .... mol
2. 14g+3(1g)= 17g/mol
34.05g/ 17g/mol=.... mol
3. 23g+1g+ 12g+ 3(16g)= 84g/mol
17.31g/ 84g/mol=.... mol
4. 6(12g)+12(1g)+6(16g)= 180g/mol
123.44g/ 180g/mol=.... mol
5. 23g+16g+1g= 40g/mol
2.2mol x 40g/mol= .... g
6. 2(35g)= 71g/mol
4.5mol x 71g/mol= .... g
7. 137g+ 2(14g)+ 6(16g)= 261g/mol
0.002mol x 261g/mol= ....g
8. 2(56g)+ 3(32g)+ 12(16g)= 400g/mol
5.4mol x 400g/mol=.... g
I cant believe i had to do this all over
Answer:
There is 52.33 grams of water produced.
Explanation:
Step 1: Data given
Mass of propane burned = 32.00 grams
Molar mass of propane = 44.1 g/mol
Oxygen is in excess
Molar mass of water = 18.02 g/mol
Step 2: The balanced equation
C3H8 + 5O2 → 4H2O + 3CO2
Step 3: Calculate moles of propane
Moles of propane = mass propane / molar mass of propane
Moles of propane = 32.00 grams / 44.1 g/mol
Moles of propane = 0.726 moles
Step 4: Calculate moles of H2O
Propane is the limiting reactant.
For 1 mol of propane consumed, we need 5 moles of O2 to produce 4 moles of H2O and 3 moles of CO2
For 0.726 moles of propane we'll have 4*0.726 = 2.904 moles of H2O
Step 5: Calculate mass of H2O
Mass of H2O = moles of H2O * molar mass of H2O
Mass of H2O = 2.904 moles * 18.02 g/mol
Mass of H2O = 52.33 grams
There is 52.33 grams of water produced.
