First you need to solve for the number of moles in each.
1 mole of anything = atomic mas of that element (g)
Carbon has an atomic mass of 12.0107
1 mole of Carbon = 12.0107 g
Iron has an atomic mass of 55.845
1 mole of Iron = 55.845 g
now convert each from grams to moles
(grams of Carbon cancel and you are left with moles of carbon)
<u>100.0 g Carbon x 1 mol Carbon </u> = 8.326 mol Carbon
12.0107 g Carbon
(grams of Iron will cancel and you are left with moles of iron)
<u>100.0 g Iron x 1 mol Iron</u> = 1.791 mol Iron
55.845 g Iron
now we can use avogadro's number to solve for the amount of atoms in each
1 mol = 6.02x10^23 atoms
<u>8.326 mol Carbonx 6.02 x 10^23 atoms </u> = 5.012 x 10^24 atoms Carbon<u>
</u> 1 mol Carbon
<u>
</u><u>1.791 mol Iron x 6.02 x 10^23 atoms </u> = 1.078 x 10^24 atoms Iron<u>
</u> 1 mol Iron
<u>
</u>5.012 x 10^24 atoms Carbon > 1.078 x 10^24 atoms Iron
<u>
</u>
Answer:
Explanation: was given the equation: Ca(OH)2 + H3PO4 --> H2O + Ca3(PO)4, which I balanced to:
3 Ca(OH)2 + 2 H3PO4 --> 6 H2O + Ca3(PO)4
I am supposed to find
a) How many grams of calcium phosphate are produced from the reaction of 25.0g of Ca(OH)2 with excess H3PO4.
b) If 0.139 moles of Ca(OH)2 react with 0.58 moles H3PO4, determine the grams of H2O produced and the grams of the excess reagent left after the reaction.
The answer is
C. condensation
Let our basis for the calculation be 1 mol of the substance.
(1 mol)(102 g/mol) = 102 g
Determine the amount of C, H, and O in mole.
C = (102 g)(0.588)(1 mol/12 g) = 4.998 mols C
H = (102 g)(0.098)(1 mol/1 g) = 9.996 mols H
O = (102 g)(0.314)(1 mol/16 g) = 2 mols O
The empirical formula of the substance is C5H10O2. The molar mass of the empirical formula is 102. This means that this is also its molecular formula.
<span>3.68 x 10²⁵ bromine atoms * 1mol/6.02*10²³ atoms=
= 61.13 mol of bromine atoms
1 mol PBr3 ----- 3 mol Br
x mol PBr3 -----61.13 mol Br
x= 1*61.13/3 = 20.4 mol PBr3.
</span>20.4 mol PBr3 <span>contain 3.68 x 10^25 bromine atoms.</span>
