Consider the balanced chemical equation: 3Sr(OH)2(aq) + 2H3PO4(aq) → Sr3(PO4)2(s) + 6H2O(l). How many moles of Sr(OH)2 remain if 0.10 mol of H3PO4 react with 0.40 mol of Sr(OH)2?
First you must determine the limiting reagent but I PRESUME that the limiting reagent must be H3PO4 (otherwise no Sr(OH)2 would be left). Then convert 0.1 mole H3PO4 to moles Sr(OH)2.
0.1 mole H3PO4 x (3 moles Sr(OH)2/2 moles H3PO4) = 0.1 x (3/2) = ?? moles H3PO4 used.
Then 0.4 - moles used = moles remaining.
0.25
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Well, let's do some "moles" arithmetic, shall we?
According to the balanced chemical equation, we need 3 moles of Sr(OH)2 for every 2 moles of H3PO4.
So, if you have 0.10 mol of H3PO4, you would need (0.10 mol / 2 mol) * 3 mol = 0.15 mol of Sr(OH)2 to react perfectly.
But if you start with 0.40 mol of Sr(OH)2, and we just calculated that you only need 0.15 mol, then there will be some molecules left over to party! But how many will remain?
To find out, we need to subtract the amount used from the amount we started with.
0.40 mol - 0.15 mol = 0.25 mol
So, there will be 0.25 mol of Sr(OH)2 that remains, ready to cause some chemical mischief!
To determine how many moles of Sr(OH)2 remain, we first need to find the limiting reactant, which is the reactant that is completely consumed in the reaction and determines the amount of product formed.
Step 1: Calculate the moles of H3PO4 and Sr(OH)2 available:
Moles of H3PO4 = 0.10 mol
Moles of Sr(OH)2 = 0.40 mol
Step 2: Set up a ratio based on the coefficients of the balanced chemical equation:
From the balanced equation: 3 moles of Sr(OH)2 react with 2 moles of H3PO4.
Step 3: Find the mole ratio of H3PO4 to Sr(OH)2:
Moles of H3PO4 to react = 2/3 * Moles of Sr(OH)2
= 2/3 * 0.40 mol
= 0.267 mol of H3PO4
Step 4: Compare the amount of H3PO4 available with the amount needed:
Since the moles of H3PO4 available (0.10 mol) are less than the moles required (0.267 mol) based on the mole ratio, H3PO4 is the limiting reactant.
Step 5: Determine the moles of Sr(OH)2 consumed:
From the balanced equation: 2 moles of H3PO4 react with 3 moles of Sr(OH)2.
Moles of Sr(OH)2 consumed = 3/2 * Moles of H3PO4
= 3/2 * 0.10 mol
= 0.15 mol
Step 6: Calculate the moles of Sr(OH)2 remaining:
Moles of Sr(OH)2 remaining = Initial moles of Sr(OH)2 - Moles of Sr(OH)2 consumed
= 0.40 mol - 0.15 mol
= 0.25 mol
Therefore, 0.25 moles of Sr(OH)2 remain if 0.10 mol of H3PO4 react with 0.40 mol of Sr(OH)2.