How Much P4S10 Can Be Made from 10g P4 and 30g S8?

[Levathian]

Homework Statement

Under appropriate conditions, the reaction of elemental phosphorus and elemental sulfur produces the compound P4S10. How much P4S10 can be prepared starting with 10.0 g of P4 and 30.0 g of S8?
Answer 35.9g

2. Step Sources
I'm following the steps from
"General Chemistry: principles and patterns v1" by Bruce Averill and Patricia Eldredge
Chapter 3.4
1. Determine the number of moles of each reactant.
2. Compare the mole ratio of the reactants with the ratio in the balanced chemical equation to determine which reactant is limiting.
3. Calculate the number of moles of product that can be obtained from the limiting reactant.
4. Convert the number of moles of product to mass of product.


3. My attempt thus far
Balanced Equation: 4P₄ + 5S₈ -> 5(P₄S₁₀)
(perhaps this is the source of my inaccuracy)

1. moles: P₄ = .081
s₈ = .1169
Limiting Reagent is P₄

2. comparing coefficients (stoichiometric moles): .081 molP₄ x (5molP₄S₁₀ / 4 mol p₄) = (.081 x 5)/4 = .100875 mol of P₄S₁₀
4. .100875 x molar mass of p4s10 = 44.83 grams.

This is however incorrect.
I get the correct answer when multiplying strait out the moles of p₄ (.0807) by the molar mass of P₄S₁₀. Is my balanced equation wrong?

Any help is appreciated!

 

[Borek]

Count atoms of any element of the LHS (left hand side) and on the RHS of the reaction equation. Are they equal?

 

[Levathian]

Count atoms of any element of the LHS (left hand side) and on the RHS of the reaction equation. Are they equal?

I'm trying to teach myself chemistry and balancing equations is still a weakness.
I got it now thank you