Answer: 24.13 g Cu
Explanation:
<u>Given for this question:</u>
M of CuO = 30 g
m of CuO = 79.5 g/mol
Number of moles of CuO = (given mass ÷ molar mass) = (30 ÷ 79.5) mol
= 0.38 mol
The max number of CuO (s) that can be produced by the reaction of excess methane can be solved with this reaction:
CuO(s) + CH4(l) ------> H2O(l) + Cu(s) + CO2(g)
The balanced equation can be obtained by placing coefficients as needed and making sure the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side
4CuO(s) + CH4(l) ----> 2H2O(l) + 4Cu(s) + CO2(g)
From the stoichiometry of the balanced equation:
4 moles of CuO gives 4 moles of Cu
1 mole of CuO gives 1 mol of Cu
0.38 mol of CuO gives 0.38 mol of Cu
Therefore, the grams of Cu that can be produced = 0.38 × molar mass of Cu
= 0.38 × 63.5 g
= 24.13 grams
Therefore, 24.13 grams of copper could be produced by the reaction of 30.0 of copper oxide with excess methane
Answer : The mass of calcium chloride is, 116.84 grams
Solution : Given,
Molar mass of calcium chloride, 
= 110.98 g/mole
Number of molecules of calcium chloride = 
As we know that,
1 mole of calcium chloride contains 
molecules of calcium chloride
or,
1 mole of calcium chloride contains 110.98 grams of calcium chloride
Or, we can say that
As, molecules of calcium chloride present in 110.98 grams of calcium chloride
So, molecules of calcium chloride present in 
of calcium chloride
Therefore, the mass of calcium chloride is, 116.84 grams
Answer: Mass is 2,37 kg
Explanation: Weight G = mg, and g = 9.81 m/s² on Earth.
m = W/g = 23.2 N / 9.81 m/s²
the overall equation for the conversation of pyruvate to acetyl COA is as below
CH3COO-COO- + NAD+ + HS-COA = ch3-COO-S -COA +NADH +CO2
The oxidation of pyruvate led to a conversation of NAD+ to NADH and produces acetyl COA and CO2
Answer:
C is the excess reactant.
Explanation:
Reaction is C + O2 --> CO2
1mol of C required to react with 1mol O2
Therefore 15 - 10 = 5moles of C will be in excess
