In this case, for the undergoing chemical reaction, we need to compute the moles of CO2 yielded by 85 g of CH4 (molar mass = 16 g/mol) and by 320 g of O2 (molar mass 32 g/mol) via the following mole-mass relationships:

$n_{CO_2}^{by\ CH_4}=85gCH_4*\frac{1molCH_4}{16gCH_4} *\frac{1molCO_2}{1molCH_4} =5.3molCO_2\\\\n_{CO_2}^{by\ O_2}=320gO_2*\frac{1molO_2}{32gO_2} *\frac{1molCO_2}{2molO_2} =5molCO_2$

Considering the 1:2:1 among CH4, O2 and CO2. Therefore, since 320 g of O2 yield the smallest amount of CO2 we infer that the limiting reactant is O2.

Best regards.

5 0

3 years ago

What is the molarity of a solution made by dissolving 18.9g of ammonium nitrate in enough water to make 855 ml of solution?

Leya [2.2K]

Answer is: the molarity of a solution is 0.276 M.

V(solution) = 855 mL ÷ 1000 mL/L.
V(solution) = 0.855 L.
m(NH₄NO₃) = 18.9 g; mass of ammonium nitrate.
M(NH₄NO₃) = 80.04 g/mol; molar mass of ammonium nitrate.
n(NH₄NO₃) = m(NH₄NO₃) ÷ M(NH₄NO₃).
n(NH₄NO₃) = 18.9 g ÷ 80.04 g/mol.
n(NH₄NO₃) = 0.236 mol; amount of substance.
c(NH₄NO₃) = n(NH₄NO₃) ÷ V(solution).
c(NH₄NO₃) = 0.236 mol ÷ 0.855 L.
c(NH₄NO₃) = 0.276 mol/L; molarity of ammonium nitrate.

7 0

3 years ago

Water contained in a reservoir is controlled by a dam. The energy of the trapped water is

Lana71 [14]

Another example of potential energy.

6 0

3 years ago