How many grams of hydrogen in 1.85moles

1 answer:

5 0

2H2(g) + O2(g) → 2H2O(1) 0 260 g 0.2068 0.180 g 2008 When 45.0 g of CH4 reacts with excess O2, the actual yield of CO2 is 118 g. What is the percent yield? CHA(g) + 2O2(g) - CO2(g) + 2H2O(g) 73.6% 67.9% 95.2% 86.4% For the reaction: 2503(g) + 790 kcal - 25(s) + 3O2(g), how many kcal are needed to form 1.5 moles O2(g)? 790 kcal 395 kcal 2370 kcal 411 kcal When 3 moles of Ny are mixed with 5 moles of H2 the limiting reactant is N2(g) + 3H2(g) - 2NH3(g) H2 NH3 ОООО H20 O N₂

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Determine the mass of 5.2 x 10 power of 21 molecules of propanol C3H7OH(l), on grams.

Alja [10]

$n(\text{C}_3\text{H}_7\text{OH}) = N(\text{C}_3\text{H}_7\text{OH}) / N_A\\ \phantom{n(\text{C}_3\text{H}_7\text{OH})} = 5.2 \times 10^{21} / (6.02 \times 10^{23})\\ \phantom{n(\text{C}_3\text{H}_7\text{OH})} = 8.6 \times 10^{-3} \; \text{mol}$

where $N_A = 6.02 \times 10^{23}\; \text{mol}^{-1}$ the Avogadro's constant that relates the number of particles to their number, in the unit moles $\text{mol}$ .

The molar mass of propanol- mass per mole propanol- can be directly deduced from its molecular formula with reference to a modern periodic table.

$M(\text{C}_3\text{H}_7\text{OH}) = \underbrace{3 \times 12.01}_{\text{carbon}} + \underbrace{8 \times 1.008}_{\text{hydrogen}} + \underbrace{1\times 16.00}_{\text{oxygen}} = 60.09 \; \text{g} \cdot \text{mol}^{-1}$

$8.6 \times 10^{-3} \; \text{mol}$ of propanol molecules would thus have a mass of $8.6 \times 10^{-3} \; \text{mol} \times 60.09 \; \text{g} \cdot \text{mol}^{-1} = 0.52 \; \text{g}$