What is the maximum number of grams of PH3 that can be formed when 41.0 g of phosphorus reacts with 22.1 g of hydrogen to form PH ? (Hint: First, find the limiting reactant)
P4(8) + 6H2(g) ->4PH3(g)
a. 45g
b. 0.43 g
c. 270 g
d. 6.8g

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answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

i believe it is a.) released into

explanation:

the answer would be b i believe