Answer:
116.3 grCO2
Explanation:
1st - we balance the equation so that it finds the same amount of elements of the product side and of the reagent side
C6H6 +15/2 O2⟶ 6CO2 +3 H2O
2nd - we calculate the limiting reagent
39.2gr C6H6*(240grO2/78grC6H6)=120 grO2
we don't have that amount of oxygen so this is the excess reagent and oxygen the limiting reagent
3rd - we use the limiting reagent to calculate the amount of CO2 in grams
105.7grO2*(264grCO2/240grO2)=116.3 grCO2
Answer:
the answer is A
Explanation:
Hopefully u get the answer right
Answer:
The molar mass of carbon
Explanation:
Before the mass (in grams) of two moles of carbon can be determined, <u>the molar mass of the element would be needed.</u>
<em>This is because the number of mole of an element is the ratio of its mass and the molar mass</em>. That is,
number of mole = mass/molar mass
Hence, the mass of elements can be obtained by making it the subject of the formular;
mass = number of mole x molar mass
<em>Therefore, the molar mass of carbon would be needed before the mass of 2 moles of the element can be determined.</em>
