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378.3g You multiply the RMM by the Concentration (mol) Mass(g)=Concentration(mol)*RMM
Given the balanced equation2C3H8O + 9O2 --> 6CO2 + 8H2OTo find the number of moles CO2 that will be produced from 0.33 mol C3H8O, we must convert from moles to moles (mol --> mol conversion).0.33 mol C3H8O * 6 molecules CO2 = 0.99 mol CO2---------- 2 molecules C3H8O
Yes, the mol is the SI unit of concentration.
C2H4 + 3 O2 --> 2 CO2 + 2 H2OSo 2.16 mol O2 will produce 1.44 mol H2O(and 1.44 mol CO2)because 3:2 = 2.16 : 1.44
Yes. CO2 has a weight of 44g/mol and O2 has a weight of 32g/mol.
Concentration is the amount of a substance in a given mass or volume of a material, expressed in percentage, mol/L, mol/kg, g/L etc.
(90 g CO2 / (12.0+2*16.0) g/mol CO2) = 2.05 mol CO2 = 2.05 mol C2.05 mol C * 12.0 g/mol C = 24.5 g C
378.3g You multiply the RMM by the Concentration (mol) Mass(g)=Concentration(mol)*RMM
2.65 mol * 64.07 g/mol = 169.79 g
1.3 mole C2H2 will produce 2.6 mole CO2, weighting 2.6(mol) x 44 (g/mol) = 114 g CO2
Given the balanced equation2C3H8O + 9O2 --> 6CO2 + 8H2OTo find the number of moles CO2 that will be produced from 0.33 mol C3H8O, we must convert from moles to moles (mol --> mol conversion).0.33 mol C3H8O * 6 molecules CO2 = 0.99 mol CO2---------- 2 molecules C3H8O
This measure is the concentration of NaCl expressed in mol/L, g/L, g/100 g (percentage).
44 grams of CO2 (1 mol CO2 from 1 mol C if enough oxygen, air, is available)
How many moles of CO2 are produced when 2.1 mol of C2H2 react?
[ 34.3(gC3H8) / 44(g/mol C3H8)] * [ 3mol CO2 / 1molC3H8 ] * 44(g/mol CO2) = 103 gram CO2
This measure is the concentration of NaCl expressed in mol/L, g/L, g/100 g (percentage).
This measure is the concentration of NaCl expressed in mol/L, g/L, g/100 g (percentage).