Question

Iron(II) sulfide has a molar mass of 87.91 g/mol. 50 grams of

Answer 1

Answer:

true

Explanation:

Answer 2

Answer:

$\boxed {\boxed {\sf False}}$

Explanation:

We want to know if 50 grams of iron (II) sulfide contains 3.011 * 10²³ molecules.  Let's calculated and check.

1. Convert Grams to Moles

We know there are 50 grams of iron (II) sulfide or FeS If we want to convert from grams to moles, we use the molar mass (mass per 1 mole). This is given to us and it is 87.91 grams per mole for this substance. Let's create a ratio.

$\frac {87.91 \ g \ FeS}{ 1 \ mol \ FeS}$

We want to convert 50 grams of FeS to moles, so we multiply by this value.

$50 \ g \ FeS *\frac {87.91 \ g \ FeS}{ 1 \ mol \ FeS}$

Flip the ratio so the grams of FeS cancel.

$50 \ g \ FeS *\frac {1 \ mol \ FeS}{ 87.91 \ g \ FeS}$

$50*\frac {1 \ mol \ FeS}{ 87.91}$

$\frac {50}{ 87.91} \ mol \ FeS$

$0.5687635081 \ mol \ FeS$

2. Convert Moles to Molecules

1 mole of any substance contains the same number of particles (atoms, molecules, formula units, etc.). This is Avogadro's Number or 6.022 *10²³. In this case, the particles are molecules of FeS. Create another ratio.

$\frac {6.022 *10^{23} \ molecules \ FeS}{ 1 \ mol \ FeS}$

We want to convert 0.5687635081 moles of FeS to molecules, so we multiply by this value.

$0.5687635081 \ mol \ FeS*\frac {6.022 *10^{23} \ molecules \ FeS}{ 1 \ mol \ FeS}$

The units of moles of FeS cancel.

$0.5687635081 *\frac {6.022 *10^{23} \ molecules \ FeS}{ 1 }$

$0.5687635081 *{6.022 *10^{23} \ molecules \ FeS}$

$3.42509385*10^{23} \ molecules \ FeS$

If we round to the thousandth place, the 0 in the tenth thousandth place tells us to leave the 5.

$3.425*10^{23} \ molecules \ FeS$

50 grams of iron (II) sulfide is approximately 3.425 * 10²³ molecules, not 3.011 *10²³ so the statement is false.