what is the maximum number of grams of PH3 that can be formed when 6.2 g of phosphorus reacts with 4.0 g of hydrogen to form PH3
Okay first lets figure out the equation
P+H2---->PH3
balance it now
we have: 2P + 3H2 --->2PH3
equation is now balanced.
let us now find the reactant in excess
2 moles of P= 3 moles of H2
2(31)g of P = 3(2)g of H2
6.2 g of P= x
x= 6.2x6/62= 0.6 g of H2
4 grams of H2 are reacting which proves that H2 is the reactant in excess since only 0.6 grams were needed.
So we use the limited reactant P to find out the mass of PH3 formed.
6.2 g of P = x
2(31)g of P= 2(31+3g) of PH3
68 g x 6.2g/62 g
= 6.8 g of PH3
I hope I am correct
The answer by mohamed is correct except your teacher may have instructed you to write P as P4. If so that will change the equation but it will not change the number of grams PH3 formed.
To determine the maximum number of grams of PH3 that can be formed, we need to calculate the limiting reagent first.
1. Start by converting the masses of both reactants to moles:
- Phosphorus (P4): 6.2 g / molar mass of P4 (123.895 g/mol) = 0.05 mol P4
- Hydrogen (H2): 4.0 g / molar mass of H2 (2.016 g/mol) = 1.99 mol H2
2. Next, determine the ratio of moles of reactants in the balanced chemical equation:
- The balanced equation for the formation of PH3 is:
P4 + 6H2 -> 4PH3
- Based on the coefficients in the balanced equation, the ratio of P4 to H2 is 1:6.
3. Now compare the calculated moles in step 1:
- The P4:H2 ratio is 1:6. Since the moles of P4 (0.05 mol) are less than the moles of H2 (1.99 mol), P4 is the limiting reagent.
4. Calculate the moles of PH3 formed using the stoichiometry from the balanced equation:
- From the balanced equation, 1 mol P4 reacts to form 4 mol PH3.
- Therefore, moles of PH3 = moles of P4 x (4/1) = 0.05 mol x 4 = 0.20 mol PH3
5. Finally, convert moles of PH3 to grams using the molar mass of PH3:
- Molar mass of PH3 (33.997 g/mol)
- Mass of PH3 = moles of PH3 x molar mass of PH3 = 0.20 mol x 33.997 g/mol ≈ 6.8 g
Therefore, the maximum number of grams of PH3 that can be formed is approximately 6.8 grams.
To find the maximum number of grams of PH3 formed, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in a chemical reaction and determines the maximum amount of product that can be formed.
First, let's write the balanced chemical equation for the reaction between phosphorus (P) and hydrogen (H2) to form phosphine (PH3):
P + 3H2 -> 2PH3
From the balanced equation, we can see that 1 mole of P reacts with 3 moles of H2 to produce 2 moles of PH3.
Next, we need to convert the given masses of reactants into moles. We can use the molar mass of each compound to do this.
The molar mass of phosphorus (P) is 31.0 g/mol, and the molar mass of hydrogen (H2) is 2.02 g/mol.
For phosphorus:
6.2 g P × (1 mol P / 31.0 g P) = 0.2 mol P
For hydrogen:
4.0 g H2 × (1 mol H2 / 2.02 g H2) = 1.98 mol H2
Now we can compare the mole ratios between P and H2 to determine the limiting reactant.
According to the balanced equation, we need 3 moles of H2 for every 1 mole of P. Therefore, the 0.2 mol of P requires 0.6 mol of H2 (0.2 mol P × 3 mol H2 / 1 mol P).
Since we have 1.98 mol of H2, which is greater than 0.6 mol required by P, then hydrogen (H2) is in excess, and phosphorus (P) is the limiting reactant.
To find the maximum number of grams of PH3 produced, we use the mole ratio between P and PH3 from the balanced equation (1 mole P produces 2 moles of PH3).
0.2 mol P × (2 mol PH3 / 1 mol P) = 0.4 mol PH3
Finally, we can convert the moles of PH3 into grams using the molar mass of PH3, which is 33.997 g/mol.
0.4 mol PH3 × (33.997 g PH3 / 1 mol PH3) = 13.5988 g PH3
Therefore, the maximum number of grams of PH3 that can be formed is approximately 13.6 grams.