how many moles of ca are needed to neutralize 8.87moles of H3po4
write the balanced reaction:
3Ca+2H3PO4 >> 3H2 + Ca3(PO4)2
looks like 3/2 * 8.87 moles of Ca
Write the equation and balance it. Using the coefficients in the balanced equation, convert moles Ca to moles H3PO4.
To determine how many moles of Ca are needed to neutralize 8.87 moles of H3PO4, we should first determine the balanced chemical equation between Ca and H3PO4.
The balanced chemical equation is:
3Ca + 2H3PO4 → Ca3(PO4)2 + 3H2O
From the balanced equation, we can see that it takes 3 moles of Ca to react with 2 moles of H3PO4. This means the molar ratio between Ca and H3PO4 is 3:2.
Now, let's calculate how many moles of Ca are needed using the molar ratio.
Moles of Ca = (moles of H3PO4 * 3) / 2
Moles of Ca = (8.87 moles * 3) / 2
Moles of Ca = 13.305 moles
Therefore, you would need approximately 13.305 moles of Ca to neutralize 8.87 moles of H3PO4.