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

• Levathian
In summary, the reaction of elemental phosphorus and elemental sulfur can produce the compound P4S10 under appropriate conditions. Starting with 10.0 g of P4 and 30.0 g of S8, 35.9 g of P4S10 can be prepared. The correct steps for solving this problem involve determining the limiting reactant, calculating the number of moles of product that can be obtained from the limiting reactant, and converting the moles of product to mass. It is important to check the balanced equation for accuracy when calculating the moles of product.
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?

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!

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

Borek said:
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

## 1. What is a limiting reagent in a chemical reaction?

A limiting reagent is the reactant that gets completely consumed in a chemical reaction, thereby limiting the amount of product that can be formed. It determines the maximum amount of product that can be obtained from a given amount of reactants.

## 2. How do you calculate the limiting reagent?

To calculate the limiting reagent, you need to first determine the moles of each reactant present in the reaction. Then, using the mole ratio between the reactants and the balanced chemical equation, you can determine which reactant will be completely consumed first. The reactant with the smaller number of moles is the limiting reagent.

## 3. Why is it important to identify the limiting reagent in a reaction?

Identifying the limiting reagent is important because it allows you to predict the maximum amount of product that can be obtained from a reaction. It also helps in determining the amount of excess reactants present in the reaction, which can be useful in cost-effective production processes.

## 4. Can a reactant be both a limiting reagent and an excess reagent?

Yes, a reactant can be both a limiting reagent and an excess reagent in a reaction. This happens when the amounts of reactants are not in the exact mole ratio as required by the balanced chemical equation. In this case, the reactant with the smaller number of moles will be the limiting reagent, and the reactant with the larger number of moles will be the excess reagent.

## 5. How does the limiting reagent affect the yield of a reaction?

The limiting reagent directly affects the yield of a reaction. The amount of product formed is limited by the amount of limiting reagent present. Therefore, if there is not enough limiting reagent, the yield of the reaction will be lower. On the other hand, if the limiting reagent is in excess, the yield of the reaction will be higher.

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