Question

1. how many grams of iron(iii) chromate are present in 1.59 moles of this compound? 171 grams. 2. how many moles of iron(iii) chromate are present in 3.42 grams of this compound? 0.0317 moles. use the references to access important values if needed for this question. an error has been detected in your answer. check for typos, miscalculations etc. before submitting your answer. submit answer retry entire group 9 more group attempts remaining

Explanation:

Step1: Determine molar - mass of iron(III) chromate

The formula of iron(III) chromate is $Fe_2(CrO_4)_3$.
The molar - mass of $Fe$ is approximately $55.85\ g/mol$, the molar - mass of $Cr$ is approximately $52.00\ g/mol$, and the molar - mass of $O$ is approximately $16.00\ g/mol$.
The molar - mass of $Fe_2(CrO_4)_3$ is:

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Step2: Solve for mass in question 1

We know that $n = 1.59\ mol$ of $Fe_2(CrO_4)_3$.
Using the formula $m=n\times M$, where $n$ is the number of moles and $M$ is the molar - mass.

$$m = 1.59\ mol\times459.7\ g/mol=1.59\times459.7 = 730.923\ g$$

Step3: Solve for moles in question 2

We know that $m = 3.42\ g$ of $Fe_2(CrO_4)_3$.
Using the formula $n=\frac{m}{M}$, where $m$ is the mass and $M$ is the molar - mass.

$$n=\frac{3.42\ g}{459.7\ g/mol}\approx0.00744\ mol$$

Answer:

1. $730.923$ grams
2. $0.00744$ moles