The amount of grams in two moles depends on the element or compound that you have two moles of. For example, the molar mass of oxygen is about 16 grams. So two moles of oxygen would be about 32 grams.
moles = weight in grams / molecular weight = 56 / 28 = 2 moles
987 grams of Ra(OH)2 is equivalent to 3,78 moles.
124,60 grams of Fe3(PO4)2 are equal to 0,35 moles.
987 grams of Ra(OH)2 is equal to 3,78 moles.
The answer is 2 moles.
The equation for the reaction between methane (CH4) and oxygen is: CH4 + 2O2 → CO2 + 2H2O From the previous information, if 25.9 grams of water vapor were formed, this corresponds to 25.9 grams ÷ 18.0 g/mol = 1.44 moles of water. This means 0.72 moles of methane reacted. If the molar mass of methane is 16.0 g/mol, then 0.72 moles of methane corresponds to 0.72 moles × 16.0 g/mol = 11.52 grams of methane reacted with oxygen.
moles = weight in grams / molecular weight = 56 / 28 = 2 moles
987 grams of Ra(OH)2 is equivalent to 3,78 moles.
987 grams of Ra(OH)2 is equal to 3,78 moles.
124,60 grams of Fe3(PO4)2 are equal to 0,35 moles.
1) First find the number of moles of methane in 27.8 g using the molar mass.See the Related Question to the left of this answer "How do you convert from grams to moles and also from moles to grams?" to do that.2) Then write the balanced reaction. Methane (CH4) reacts with oxygen (O2) to form carbon dioxide (CO2) and water (H2O). See the related question "How do you balance a chemical reaction?" to do that.3) That will tell you the ratio of moles of methane to moles of oxygen (it will be 2 to 1). So from Part 1, multiply the number of moles of methane by 2 to get moles of oxygen. Then, use the Ideal Gas Law to find out how many liters that will take up at STP. Use the Related Question link "How do you solve Ideal Gas Law problems?" to do that.
The answer is 2 moles.
There are 29/14, or just over 2 moles of nitrogen in 19 grams.
The answer is 224,141 grams oxygen.
2.,7 moles of glucose have 486,432 g.
When methane undergoes complete combustion, the equation for the reaction is CH4 + 2 O2 -> CO2 + 2 H2O. This shows that the number of moles of carbon dioxide formed are the same as the number of moles of methane reacted, so that 14 moles of carbon dioxide will be formed from 14 moles of methane.
To determine the grams of CO2 produced from 2.5 moles of O2, we first need to consider the balanced chemical equation for the combustion of a hydrocarbon (e.g., methane): CH4 + 2O2 → CO2 + 2H2O. From this equation, 2 moles of O2 produce 1 mole of CO2. Therefore, 2.5 moles of O2 would produce 1.25 moles of CO2. Since the molar mass of CO2 is approximately 44 grams/mol, 1.25 moles of CO2 corresponds to 55 grams (1.25 moles × 44 g/mol).