Answer:
94.57 g H2O
Explanation:
32.0 g H2
Molar Mass H2: 2.01 g/mol
84.0 g O2
Molar Mass O2: 32.00 g/mol
Molar Mass H2O: 18.02 g/mol
First balance the eqution:
2H2 + O2 = 2H2O
Find the amount of moles that both 32.0 g of H2 and 84.0 g of O2 can produce using the molar masses:
32.0 g H2 x ((1 mole H2)/ (2.015 g H2)) = 15.88 moles H2
84.0 g O2 x ((1 mole O2)/(32.00 g O2)) = 2.63 moles O2
Now you can find the amount of grams of H2O each reactant will produce:
**keep in mind of mole ratios from balanced equation**
15.88 moles H2 x ((2 moles H2O)/(2 moles H2)) x ((18.015 g H2O)/(1 mole H2O) =286.07 g H2O
2.63 moles O2 x ((2 moles H2O)/(1 mole O2)) x ((18.015 g H2O)/(1 mole H2O) = 94.76 g H2O
94.76 is the final answer because this is the limiting reactant, meaning it produces less product than the H2 so it limits the reaction from producing anymore product from the amount calculated.
If two nonmetals are bonded together. So look at your periodic table of elements and decide if the two elements that are being bonded are nonmetals or not.
Explanation:
According to the law of conservation of mass, mass can neither be created nor it can be destroyed. But it can be simply transformed from one form to another.
Therefore, when 
is added to NaCl then the compound formed will have same mass as that of reactants.
Total mass of reactants is (169.87 + 58.44) g/mol = 228.31 g/mol
Total mass of products is (143.32 + 84.99) g/mol = 228.31 g/mol
Thus, we can conclude that mass of the new mixture will stay the same.
- Standard reduction potential of Ag/Ag⁺ is 0.80 v and that of Cu⁺²(aq)/Cu⁰ is +0.34 V.
- The couple with a greater value of standard reduction potential will oxidize the reduced form of the other couple.
Ag⁺ will be reduced to Ag(s) and Cu⁰ will be oxidized to Cu²⁺
Anode reaction: Cu⁰(s) → Cu²⁺ + 2 e⁻ E⁰ = +0.34 V
Cathode reaction: Ag⁺(aq) + e → Ag(s) E⁰ = +0.80 V
Cell reaction: Cu⁰(s) + 2 Ag⁺(aq) → Cu⁺²(aq) + 2 Ag⁰(s)
E⁰ cell = E⁰ cathode + E⁰ anode
= 0.80 + (-0.34) = + 0.46 V
