Question

3fecl₂ + 2na₃po₄ → fe₃(po₄)₂ + 6nacl

1. g na₃po₄ are used during the reaction. the reaction began with 41 g of na₃po₄. what mass of excess na₃po₄ remains when this reaction has gone to completion?

? g na₃po₄

Explanation:

Step1: Identify the given amounts

We know the initial mass of $\ce{Na3PO4}$ is $41\ \text{g}$ and the mass of $\ce{Na3PO4}$ used in the reaction is $20\ \text{g}$.

Step2: Calculate the excess mass

To find the mass of excess $\ce{Na3PO4}$, we subtract the mass used from the initial mass. The formula for excess mass ($m_{\text{excess}}$) is:
$$m_{\text{excess}} = m_{\text{initial}} - m_{\text{used}}$$
Substituting the given values:
$$m_{\text{excess}} = 41\ \text{g} - 20\ \text{g}$$

Answer:

$21\ \text{g}$