Which statement is true for the reaction represented by this equation?CH4+2O2→CO2+2H2OGroup of answer choices1 mole of CH4 is required to react with 4 moles of O2.1 gram of CH4 is required to react with 4 grams of O2.1 mole of CH4 is required to react with 2 moles of O2.1 gram of CH4 is required to react with 2 grams of O2.

16.0 grams of H2 react with 64.0 grams of O2 according to the following equation:2H2(g) + O2(g) → 2H2O(l)Determine which of the following species will be present in the reaction mixture at the end of the reaction. MH = 1.01 g mol-1MO = 16.00 g mol-1H2, H2O, and O2O2, H2OH2, O2H2, H2O

In the chemical reaction CH4 + 2O2 → CO2 + 2H2O, there are reactants on the left of the equation and products on the right. If you count the atoms in the reactants, there are Group of answer choices1 carbon atom, 4 hydrogen atoms, and 4 oxygen atoms4 carbon atom, 4 hydrogen atoms, and 2 oxygen atoms4 carbon atom, 4 hydrogen atoms, and 4 oxygen atoms1 carbon atom, 4 hydrogen atoms, and 2 oxygen atoms

Consider the reaction shown in the following balanced equation:2 Cu2S (s) + 3 O2 (g) → 2 Cu2O (s) + 2 SO2 (g)You have 1.00 mole each of Cu2S and O2.How many moles of Cu2O can be produced if 1.00 mole of Cu2S reacts? Answer 1 Question 1How many moles of Cu2O can be produced if 1.00 mole of O2 reacts? Answer 2 Question 1The amount of Cu2O that can be produced is smaller from 1.00 mole of Answer 3 Question 1 than the other reagent. Therefore, Answer 4 Question 1 is the limiting reagent and Answer 5 Question 1 is the reagent in excess when 1.00 mole of each is available.  Consider instead that you have 1.00 g each of Cu2S (molar mass of Cu2S is 159 g mol−1) and O2 (molar mass of O2 is 32.0 g mol−1).1.00 g of Cu2S is Answer 6 Question 1 moles of Cu2S and would produce Answer 7 Question 1 moles of Cu2O if all the Cu2S reacts. 1.00 g of O2 is Answer 8 Question 1 moles of O2 and would produce Answer 9 Question 1 moles of Cu2O if all the O2 reacts.The amount of Cu2O that can be produced is smaller from 1.00 g of Answer 10 Question 1 than the other reagent. Therefore, Answer 11 Question 1 is the limiting reagent and Answer 12 Question 1 is the reagent in excess when 1.00 g of each is available.

Considering this balanced chemical equation, how many grams of HgO will be produced when 44 g of Hg react with excess O2?2Hg (l) + O2 (g)→ 2HgO (s) Group of answer choices48 g96 g44 g28 g

Balance the following equation: ___NH3 + ___O2 → ___NO2 + ___H2O The balanced equation shows that 1.00 mole of NH3 requires ___ mole(s) of O2Group of answer choices3.51.251.751.330.57